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(9)
102
2) with n = 1. We also satisfy in part condition
we satisfy conditions 1) and
4):
specifically, the function c;l>1(Z') is regular in the disk
Iz' I < 1
103
§3. VARIATION OF UNIVALENT FUNCTIONS
III. REALIZATION OF CONFORMAL MAPPING
la"l~ (n~I)~' Ib"I~(n-:W'
since the
n=l, 2, ... ,
bracketed expression in (7) has no negative powers of z' and 1 (0) = O. (We note that the function ¢ 1 (z') is precisely determined up to a purely imaginary constant
and if 00
term by the conditions listed.)
CD
Let us suppose now that the functions
¢.)z'),
I z'l
n (0)
=
O. Furthermore, condi
¢n (z')' (Here, and in general, ¢" (z') is determined up to a purely imaginary constant term.) If we choose ¢" (z ') succes 1) and
~ ,.~~Bm U2 + 312 +... )= (n ~~.
( 11)
k=-oo
I z' I < 1
2) are satisfied for
sively, for n = 1, 2,' ", according to the rule given above, conditions 1) and 2)
where C = 8 (Z-2 +
r
2
functions ¢ (A)k, we get the conclusion of the lemma. Let us make a convention on notation. For any Laurent series x(z) =
1': =
-00
c n z " we define
(X)
IIx(z)ll= ~ Ic"llzl"· n=-OO
We still need to investigate the convergence of the series appearing in condi tion 3), on which the satisfaction of condition 4) depends. To do this, we need to prove the
Lemma. Suppose that the function ¢(A) has the expansion ¢(A) = l':;'=l a the coefficients in which satisfy the inequalities Ian I :::; I/(n + 1)2 for n = 1,2,.... Suppose that the kth power of this function has the expansion [¢ (A)] k = 1': =1 a~k)An
When such majorizing series converge, they possess the obvious properties
II XI (z) +"1.2 (z) II ~ II XI (z) II + II X2 (z) II, II Xl (z) X2 (z) II ~ II Xl (z) 11·11 X2 (z) II·
"A n
where k is a positive integer. Then the coefficients in this last expansion satisfy the inequalities la~k)1 :::; Ck/(n + 1)2, n = 1, 2,"', where C is an absolute constant. Proof. If the coefficients in the expansions of the two functions
(j' (A) = ~ a"A", n=1
satisfy the inequalities
00
~(A)= ~b"A" ,,=1
,
+ •.. ). If we apply this inequality successively to the
will be completely satisfied and condition 4) will be formally satisfied.
co
22 ~)"
1 I 1]
(:+:)' + ... + Int']' .(n-r~1+2)'
+3'·
then the bracketed expression in (6) will contain no negative powers of z' and, tions
1
(10)
(n)
(~)Z'k+ _~ Z'k' ~
"~~-oo
[I I
00
k=-oo
hence, c;l>n (z') will be regular in
(A) HA)= ~ ("All
(12)
Let us investigate the convergence of the series ~:~' =1 A ¢)z'). Let us show V
that the inequalities
,
DM,-l
1I(j'.(z)II~('1+I)'-' v=l, 2, ... , hold in the annulus r :::; I z'1
::;
llr for suitably chosen constants D and M.
To prove the validity of (13) for v of the quantity 4
1\ ¢1 (z')1\
(13)
=
1, we need only take for D the maximum
in the annulus r:::;
I z I : :; I
Ilr.. Let us seek, by choos
ing M suitably, to satisfy inequality (13) for all v. Let us suppose that inequal ity (13) holds for v = 1, 2,"', n - 1 and let us find a bound for
II¢" (z') II·
To
104
III. REALIZATION OF CONFORMAL MAPPING
§3. VARIATION OF UNIVALENT FUNCTIONS
105
do this, we consider the function (10), wbich we denote by r(z'). We define ~
F (z, ),) =
II'tn(z') II ~ B
AkIZ k).,./.
(4)
k=-oo.oo 1=0,00
This series converges for bave
II 'tn (z') ,,~ I z't' (z') 1
r
< I z I < 1 and
I
II
I Akll r
k
~
. '1.1 dAn ~ I Akll k,l 'I. -
~
,
Ik)"le
n! dAn
v7:1 (v+I)"
I
Ae
~I
v=1
n
00,
l+j
J=O
k, I
(if we replace >.AI with A\ where R = max
fl. I
I
is a regular function for r:S
ll
=r 1/z'
'I.
nldAn
Al (
A'
'n.
In particular,
"
A=O
-I ~
(15)
(n)
~ le k
k,=-oo
f' (z ')/1.
, Ik
DMII-l
liz I ~2(n+I)2'
(17)
Furthermore, since the majorant series of the second sum in (11) does not exceed the majorant series of the first sum, for
Iz 'I < 1,
it follows that (17)
yields the inequality
DMn-l
I Cl'n (z') 1\
"
Y~_)J I,=o-(n-I)ldAII-I __1_ d n- I (~_A_V )J I """,(-v+1)2 '=0
on the circle
Iz' \ =
'I.,
(13) holds for all v in the annulus r:S
I.-:
n-I
v)1 = ~ "UP' c.
= tAl'¢)z
~ ('I.
+ I)'
r and hence in the annulus r:S
For M chosen independently of
.=1
is the coefficient of An -I in the expansion
\• =1
Tberefore, this last sum in (6)
DMn-l
..0.1('1+1)2
( " A ~ ~+ 1)1
IAI < Ao.
II'tn(z·)II~~.
~V (-v +A'1)2 )JI
n-I
.=1
and
M (tbough independent of 'I.), less than DMn- /2(n + 1)2. Thus, for such M on the circle I z' I = r, we have
we may assume tbat l + j ~ 2 and, obviously, that l:s n. Furthermore, since
-l_~
Iz I < 1
M since tbe series (14)
1
are equal to O. Therefore, when summing with respect to k, l, and j, n-I
DC
ber in (16) is, for sufficiently large
Now, the terms with j = 0 and j = 1 for 1=0 and the terms with j = 0 for
l? 1, l';'
I kl + 1)9 rkM-It---xr
with large values of M is of the order of 1/M 2 • Consequently, the right-hand mem
--.w- v~ = I (V+I)2/ bO (
(16) I
2, this sum will be equal to
I z' I = r,
v=1
tz'
~
~ I Akll (t
A=O
k Ii Di I d n I j!MF 'I.! lii.." A
+ l)~ rkM-I- J I k 11~)JCj JI
where N is independent of z' . If in the last sum we drop tbe requirement that
A=O
n-I
=B.! I A k/l rkM l! I
J
and this series converges for sufficiently large values of
IklD ~
n n-I I d I Ak/lr M n!dA nA e
~ I Ak/l (l l+J~2
n-I k
I k JJDiGi "I ( J. n-I+W
~
NMn
('1.+ I)'
k, l.
n-l Ikl.E AVII'Pv(Z')llj
n-I AVDM V- 1 Ikl~ _ _
~I
=B!I
Iz
I·i
IAk/l(l+ l)jrkMII-I.i" Ik/iDJCi J![(l+I)(n-l-fl)J2
I A=O
1, we obtain on the circle
1 d n_
I rkM n
kl,
'I' (z') r
kn
"
1 dn
Then, by (13) for v = 1, 2, .. "
!
k.l~J
k, 1
~ I z't' (z')
II 'tn(Z), II ~B
=8!
0 0 and for sufficiently small A> 0, the function (20)
k=1
'"
~_I
k zf' (z) f (z) -
-.'-
=1(W'-W')(I+A),+A
"
has only simple poles in the disk \ z I < 1 at the points z k' k
(20)
S (z) -
w* (W'') I
=
k=1
Ak
w-plane.
I w* (w') -
Akf (Z)2
...
..!!....!fL _ '\'1 A f (z)2
k=1
=
fez k), k
The theorem that we have proved is applicable to this function. Since the function
m
p, k
=
k=i
sider first the function
=
such that w k
m
Let us give some examples of the application of this theorem. Let us con
WI'"
I ••• , Zm are points of the disk \ z
•
w*=w+AW(A+
and that it is normalized by\the con I
and univalent in some annulus r
(7) and (9)
IZ \ < 1
1. On the other hand, if the points wI"", w m all lie in tHe inter
ior of the domain B, then, for sufficiently small A> 0, the function (21) 1s regular
(19)
y=O
where «1>0 (z ')
107
§3. VARIATION OF UNIVALENT FUNCTIONS
III. REALIZATION OF CONFORMAL MAPPING
106
k
(
)II--.!.!L Z - Zk
f(Zk) )i Zk~ zk!' (Zk) I - ZkZ
+ 0 (Ai).
(24)
lei__"lII{.lIIIlnll!.~'lfl_.ll1IIl"
~-------
mi..... r-FI1.-!WiIIMliIilU~~ _ _~ ~ ~ ! Q l N ~ ~ ~ i l i i : _ I I I _ " ' £ _ ~ ~ ~ _ 9 M i (
III. REALIZATION OF CONFORMAL MAPPING
108
§3. VARIATION OF UNIVALENT FUNCTIONS
f(z), then, for arbitrary
w
From these formulas we can easily obtain the corresponding variational
trary A and for sufficiently small A> 0, the function
wk either lying inside
formulas for the function
m
II
F(~)=~+l1o+~1 + ... , which is univalent and regular in the domain ( =
B or external to B, for arbi
Formulas (21) and (24) are the desired variational formulas. I)
=
00
I (-I > 1
(W-ak)
w*=w+AAwk;:1
except for the pole
(27)
II
aod which does not vaoish in that domain. To do this, it is sufficient to
(W-wk)
k=1
apply (21) and (24) to the function
will, as was shown above, map the portion of some neighborhood of the boundary ; of B that lies in B onto a doubly connected domain and it will map the boundary
I
F(;)
!(Z) =
itself into the boundary of some simply connected domain BA that includes w*
=
O.
Furthermore, the doubly connected domain referred to above will lie in BA' Since
and set
.F. (~) = (We note that F * (()
to
in
(I I ) = ~ + 116 +': It[ +
the points al"", am do not lie in B and are mapped by (27) into themselves, , . 'll'
it follows that, for sufficiently small A, they will lie outside BA. Therefore, the function w* = (z), which maps the domain 1 z I < 1 onto the domain B A does
I. C
r
I (I > 1.)
oot assume in
Simple transformations lead to the following formulas:
F* F ({) = F (~)
*
+ A"'"~ A m
m
l: A
(q=F(~) +A k=1
k
k
+ AF (~) k~1 '\1 A (--f k CkF' (C
F~~;~(~k + O(A~),
k
Wk
of the domain
~
) )
k
_I=--+- 0 (A ~), 1 --- C:.C
under the mapping by the function w
k=1 m
II A o= A,
(26)
=
and that does not assume in
=
A k = A -m=--- II (Wk-W.,) .,~I
~=;:'k
F«().
Then, we find the formula for
[* (z): m
II
f«() tha t is regular and univalent in
Iz I < 1
certain given finite values a I •...
IT
am in
I/(I- \zol> .
-> ""
the rh!ht-hand member approaches
Consequently, the upper bound for arg [f(z)/z]
given by inequality (6) is tbe least upper bound. In tbe same way we can sbow
1that the
Consequently, it represents that branch of tbis multivalued function thar approaches
lower bound for arg [{(z)/z] given by inequality (6) is the greatest lower
1bound.
Since the function
O.
Inequality (6), which we have proved for tbe class S I, holds also for the
g(\.,)=
Let us show that the bounds for arg [f(z)/z] given by inequality (6) are
:s t
=~=""~"'~. '~.=""""""",.,,,,,,,,,,,,,==.=---~-=;=,==~,,,~~~~,,,,,,~, ,,,._~-.,,,,
§l.
IV. EXTREMAL QUESTIONS
114
I+z(
.
()
Consequently, for functions f(z) in the class S, we have l )
O. The bounds given by inequality
-->
2 I Z 1 Z
(= - z, we again obtain (6). atg
f' (z).
f' (z).
+
ot -
from which we get
-:;.1
iJt
for
I z !;;::: Y2'
I
12
(13)
I' _
(I +2kl- kIP) _ (I - k/)2
01 (z, t)
-
iJz
2~
«1- k/)2)
-
II-
,_
d t argl -
2S' «(1- kf)') 11
1..e12
-
dill
I""
z
'
in the class
kl If
5, we have
o
for every p S
,,,.,,
p
1/J2,.
The sharpness of the bound (13) (for arbitrary z in that is continuous in (0 St
~1 sin2arcsinlfl=2IfIVI-I/I~ I
I
1/\ ~Y2'
for
J
II I - YI_1/12
for
4d
II I
1-III (1-//1
•
1/1~y'2'
for
< 00)
I argj' (z) I ~ \ y 4d III
.:1
«J -
- 2 \I kl)2)
for
1-1/1 2
-I
II I
1/1-=Y2'
III ~ Y2' Iz I S 1/J2,
..3M. -, =
.
Sin
-1/\
=
arg(1- kl)
o
for
1/\~Y2
I
=4 arc sin I zl,
for arbitrary t in the interval 0
I
for
\/l~~,r-' ,2
for
1/1;;:::Y2
I
\
largj'(z)l~ .) YI_1/12+
\/1 ~ Y2'
or, what amounts to the same thing, the condition
-Y2
S t < 00 and given z
I
(14)
= z 0 in the disk \ z
I < 1.
From (14), we obtain by simple calculations
III (\/\- tV 1 - 1112) 4d
I
for
.
I
2)
I V I -If [2
-1-I=k/1 2 = sm2arg(l-kl)=
I z I ~ 1/y"i, C
in fact) can be
for which f(z, t) defined by equation (2) and the
Iz I
l'2
Iz I < 1
condition f(z, 0) = z satisfies the condition
\
~ _I_
Integrating this with respect to t from 0 to 00, we obtain for
~
rJ.= arc cos p,
seen by showing that there exists a function k (t) of absolute value equal to 1
II- k/1
2d
is obvious
)
I arg/' (p).[ =4 arcsin
13 «1-kf)2) I =jsin2arg(l-kl)1 2
Idt arg I I ~
I z lSI/Vi
.
C I elC1z I(z)= ~ (I~. ia~\8 dz = z+ C2 Z2 + ... ,
But
Therefore,
for
from the fact that, for a function
and, consequently,
and, for
I z I ~ y2 '
I arg f' (z)1
That this inequality gives a sharp bound on
I 2kl - k 2p (1-kl)2'
01' _ _ I'
.
for
In an equally simple manner, we can find a bound for
Specifically, from (2) we have
oarg/' =
I
f 4 arc sin I z \ I arg j' (z) I ~ ~l 7t + log I _I
(12) are also sharp because, when we apply (12) to the function (11) at the point
A hound for arg
I Z 12
(argl (z)\~7t+log I-Iz/!'
0
zl'(z) C C I(Q-/(z) arg I(z) =.) d arg V' (z) (1 _I z I') =.) d arg ~i- , o z that is, the branch that approaches 0 as z
115
that is,
C+Z) I (- --I z
-z
ROTATION THEOREMS
Izi
\
r
_ Y2
II I 1/1(1-1//2 ) ' 2d
kl=
I±Y21./12-1
\
')
.1=t-Y21/1 2-1 2
-l
1
for
1/1~y:r'
for
III ~ y~(
I
1) Inequality (13) was proved by the author (see Goluzin [1936] ). That it represents a sharp bound for z > IIV'2 was shown by Bazilevi'C [1936a].,
I I
116
§ 1.
IV. EXTREMAL QUESTIONS
±.
Here, we can take either of the signs terms of
I fl,
On the basis of this equation, we have for arbitrary real
Now that we have found kf expressed in
pendent of t,
let us substitute the result into (4) and (5). Then, from (4) we obtain
as a continuous monotonic function of I fl and, consequently, I fl as a continu ous and positive function of t. From (5) we then have a representation of
ffi
t
i l
I' + log -.-I ,-._
This shows that the upper bound for arg least upper bound for arbitrary z in
1
f' (z)
for
I(z)
'U vI
I I Zo I;;::: Y2'
Since given by inequalities (13) is the
I
I
I!)
we
(15)
I
z 1 < 1. Therefore, by shifting from f(z) to the function F(')
const, we immediately obtain sharp bounds in the class
I arg F' (C) I ~ -log ( I -
I.
=
I 'I
f' /
\ log F' (C) I ~-Iog
< 1.
(19)
We note that we can show
e.
F (0 € I. in the domain
(1 - I elll)'
(20)
Furthermore, this inequality, from which one can get the inequality
dif 1 -
:1".
1/
rfr~ IF'(C) I ~
I~I_'
constituting the distortion theorem for the class
(21)
Ie 12
I
olog 7 2kl -a-t- = - ( l - k f ) I '
from which we get
(18)
'> 1 we have the sharp inequality
(16)
f,
S and all z in \ z I
In the same way we can show that for the functions
We note that, by modifying slightly the preceding derivations, we can some times obtain stronger results. For example, if we divide equation (2) by ferentiate with respect to z, and then divide by f, we obtain
(17)
by a procedure like the one used above that inequality (17) is sharp for arbitrary
F '(,):
n-r)
----=:
~) + log (1 -I zit) \ ~ log ~ + \: i
which also holds for all fez) €
l/f(1/ ,) +
for arg
\ log I
,~~
which gives sharp bounds for arg [z2f ' (z)/f(z)2] at an arbitrary point z in
(e-IO log zl'I (z)(Z)) ,,;:::: log II + Iz I -I z I'
apply (17) to the function (11) with' = -z, we obtain another inequality: 1)
:l
9
2d II I - I I + Iz I ~ 1-1/1 1 - og 1-1 z I
I
~rJ
If
~-log(1-lzl),
-..::
e is arbitnlry, this is equivalent to the inequality zl' (z) I I + 1z I log I(z) ~log l-Iz\'
We note that, reasoning as above, we can prove the inequality zl/' (z)
from 0 to 00,
S we have in \ z I < 1
Consequently, for all fez) €
z I < 1. In the same way we can show that
t
I ZI
the lower bound given by inequalities (13) is the greatest lower bound.
arg I(z)'
2d III . (e-10 d log I') 7 .:;:;:; - I=T7T1
ffi (e- iO log zl' (Z)) -e \
I for I zol~'Y2' ZO
11:
which we assume inde
we obtain
For the function (1) constructed from this k(t), we have, as we can see from
the proof of inequalities (13),
argj' (zo) =
e,
If we integrate both sides of this last inequality with respect to
arg [f(zo, t)/zo] as a function of t. Knowing arg kf, we find arg k(t) and thus k (t) itself.
4arc sin Zo
117
ROTATION THEOREMS
I.,
will be derived later by a
different procedure.
olog 71'1
I-o-t-
"
21 II
= II-kill'
Eliminating dt here and in equation (4), we arrive at the equation
d log f'
I
I
I= -
(We give the following definition: suppose that a functional II: is defined
~. on a certain class of functions g(z). The set D of values w that the functional w = II: assumes on the given class of functions is called the range of values of the functional II: in that class of functions.
2d I I
I .
I-I/I~
1) Inequalities (18) and (19) were obtained by Grunsky [1932] by a different procedure.
_ u __ ._._"
•.
_ _ u •• _ . _ _
~ _ . _ .
-"
~ _ - ~ - - - - . - - - - - - ~ - "
118
-----._,.-:--
~
..-- _
.
_ _ .. _ .
.."',
~~.
' _ ._ _ '
._"'
' _ . _•• _
.•
...,,,,,. .-
~!
-
._.. .'... __... '~
... "..".. ..... _- . ..
.-'- _"_.,,
.,.,~~._
,,_~
~
"' v_'-"
. . "'-
_,._~_':.~
As will be shown in §3, p. 133, inequality (20) determines the range of
2.
_~,_
. __
.~.".~._
_~_
,_ _._
...
_~~
' ' • .._."' __.,_
.. __._",._.,__.
__.n
~_~
__..
._, .............
""""'"'--~....,~."....".........""'~.-.~ _'.~_",_·~_.~_"
_,.~_,."
~.~~:~~
~_
. ,.'
__.
.
_.
.n·._._..'_'._.._._·.. .. .
."",."",..,....,~""""'''~~-.,..
~'''._._.~_·,.~_
...
!!'!!.. _ ~ . _
,'~".,..,..
~'''_''
_·_'
....
"".-''''''"'
where we set
By the same reasoning, we can
C, =
show that inequalities (I8) and (19) determine the ranges of values of
I Zv'
F (C) = f
I
(-H
j
F (C) E ~.
log [z['(z)/[(z)] and log [[(z)lz] respectively in the class S. For more on this We introduce the expansions
point, see also Lebedev [1955aL)
co
§2. Sharpening of the distortion theorems
k.I=1
k (t)/(z, t) ~ k l-k(t)/(z,t)= ~bk(t)Z,
present sharpened forms of certain inequalities constituting distortion theorems for the classes Sand
2.
x , _ 1, denote real numbers such that 2 v v I =1 a v v IX v'v is a pos itive quadratic form. Then, [or arbitrary C::v ' v = 1, 2, ... , 'n, the' inequalities
the expression ([v-[v,)/(zv-zv') in (I) should
II II _ ---ln
f' (zv)'
If we integrate equations (1) with respect to t from 0 to "" and keep (1) and,
,
I
V
'I \ _
\l, lC'I>I, ~
log F(C)-F(C')= y y,
'11'=1
I
["V, v'
1- cve"
1'I
> 1, where we should understand the [actor IF' (C::) I in the second product when C::V = C::v ' .
hold throughout the domain
Proof. Let us first prove inequalities (6) for the functions F(C::) where [(z) €
S I . In this case, if we set
=
....
......
.,."..,.-."."..,.
'-'._"~~~_~'_"~.~_ 1.
~ IvI.,logF(C.)-F(C,)
C -c ,
!
In an analogous manner, we can prove the more general
Theorem 2. Let n denote a positive integer. For v = 1, "', n, let Y)I denote arbitrary complex numbers and let (IJ denote complex numbers in the domain 1(' > 1. Then
",'=1
n
co
~2~
determines the range of values for the summation on the left •
v'=]
quently they remain valid for the entire class theorem.
F (C,,)
As will be shown in §3, inequality (7) is sharp for arbitrary YIJ and (IJ' and
proved for the functions F (C) E: ~ represented in the form F{(,} = 1/{(110, where !(z) E: 5', obviously remain valid for arbitrary functions FCC) E: I that can be
II{Ol ()
.=1
completes the proof of the theorem.
We then immediately obtain inequalities (6). But these inequalities, which were
represented in the form F (()=
J
From the second of formulas (2) applied to the last integral, this yields (7) and
n
~ a v, v,XvX" ~ 0, ~ a v, v' YvYv' ~ 0.
.., '11'=1
(C) -
v
We multiply these formulas by the numbers alJ IJ' and sum each of them with with respect to v, v' = 1, ... , n. Then, rememberi'ng that for numbers a)l)l' that , satisfy the conditions of the theorem, we have n
'\1 kf)9 ."-IVI-kf, dt
\' ( 0
Cv -C·"
',v'=1
11 - ~ I+ 4SY V"~ dt. C,C.,
kf.,
',v'=1
0::>
P
kf
1.IV'I_kj, l_kf.,dt=-2
and, consequently,
co
4
n o o n
r" =-2 J ko 00
o
log
F(C;)-F(C,,)
"I'~~l
Then, by adding and subtracting, we arrive at the following formulas
F (C,) -
121
§2. SHARPENING OF DISTORTION THEOREMS
! l;r.,IOg(l----l-) C.C
v,.' = 1
v'
(7)
F(C)-F(C')
C-C'
,
that approaches zero when at least one of the POintS {;, ~
,
approaches
For {;,. ~' , this quantity becomes equal to log F' (0 and then it can be understood as the value of that branch of the multivalued funcrion thaI vanishes at {; = 00.
00.
122
~ (If
y-
with equality holding for the same function F«(). n -
TI'
1
I
._,
from (9) for n
1
V 'v,v' log 11 --_-) Y"r.,IOg(l---;:=;-). ~ \ C'lI Cv' ",,,'~1 .... "vCv' v,'Y'=l
we write inequality (8) for F 2 «() €
123
§2. SHARPENING OF DISTo.RTION THEORI;idS
IV. EXTREMAL QUESTIONS
(8)
=
n'
=
I and
1(11 = I({ I =
(Inequality (ll) also follows
p > 1.)
The applications of formulas (4) and (5) are effectively used. Theorem 4. Suppose that fez)
~2' replace nand n' with 2n and 2n' ,
P (Q = _1_
and set YlI+v=-Y v and (lI+V=-(V for v=I,''',n and Y~'+v =Y~ and (~'+vo= )', , . -':.v for JI = 1,"', n , we obtain the
t(i),
ICI> land'Jog
.
=
~';' =ICllZll €
S. If we set
F (C) - F (t')
C'
c-
00
Corollary. Let nand n' denote positive integers. Let Y v for v = 1,"', n and y~, for v' = 1"", n' denote complex numbers. Let (v' for v = 1,"', n
and (~, for v'= 1,"', n', denote complex numbers in the domain
1(1) 1.
- k,l=l 2: ak,IC-kC'-I, ICI>I, le'l>l
Let
where the ak, I are rational functions of the coefficients c k, then, for arbitrary
F 2 denote a function belonging to ~2' Then,
v "-V'v"'.1og [F (C v) "v= 1.'=1 F (C v) + F
values of the complex variables Xk (where k
!
+ C:'1
" v (C.') C. - C'
i
I
CvCv' + 1 " ,-, CvCv' + I ' v ," log C - _ ( "- ,v,v' log ,-, . " v'=1 vC., v, v'=1 C,-C v' V
~
~
(9)
z/(l ±z)2. Proof. We can confine ourselves to functions fez) € tions (5). Therefore,
We mention some special cases of Theorem 3. In the first place, if n
F (C') 1,,;:::: -log
F (C) -
C-C'
~
with equality holding for the function F «() = ( + e i
n' = 1
Second, if n = n = 2, Yl we obtain the inequality
(1 _l-) pi , ag €~,
=
,
Y2 '" 1, Y2 '"
k.I=1
>:-1,
n
where a is a real
':.2 =-':.\1
and
1
+
k
+ pi"1
001
't}k (t) X k dt = 2 ~ k!t=lbk (t) b l (t)Xkx1dt.
0
k=1
l
2
n
1
00 n
=
1, 2," "
=
(13)
corresponding to them are real. Therefore, equality holds in the =
1,2,·..
•.. , n, are real. This completes the proof of the theorem.
which, if F«() is an odd function, yields the sharp inequality of Bazilevi'C [1951] and Milin (see Lebedev and Milin [1951]):
1"(~~ Ill·
~:,';
Schiffer [1948] proved the sharpness of inequalities (12) for arbitrary complex values of xk'
'",(
As a generalization of inequality (12), let us prove, under the same condi
1
I~ log 1+ 71; I - pi
=
± 1 for 0 < t < "'", so that the functions bk (t) for
conditional inequality (13) for these functions when the values of x k' k
p2
(F(C)-F(C')) (C+C') C')
k, 1=1
k,~lak, lXkX 11 ~ 2 ~
z/(1 ± z)2, we have k (t)
1-
I log (F (C) + F (t')) (t -
y
0
On the basis of (5), this yields inequalities (12). In the case of functions fez)
':.2 =-':.1,
(F (C) - F (C'» (F (~P-{-C'») (C C')21 log (F (C) - F (- C'» (F (- C) - F (C')) (C - C')2 ~ 2 log - - 1
I
oon
~ bk (t) bi (t) Xkxl dt = -~~ (2: bk (t) x k dt
and, consequently,
(10)
, ) ' ) ' ) ' , ) ' ,
Yl
oon
TI
2: ak, IXkX, =.- 2 ~
\
constant (Goluzin [1947 ]).
.,
=
then (8) gives the sharp inequality (the subscripts are
II og
5', In this case, the
ak,l have an integral representation (4) with the functions bk(t) subject to condi
started with the area principle.)
I = p > 1,
x:
I
Theorem 3 and its corollary were first obtained by Lebedev [1951], who
1(11
1, 2,··· , n with n ~ 1),
ak,lxkxI ~! I II . (12) k,I=1 k=1 ,r- - - - - . : ."""". . . - - " . , - - - - -,-, -- 11 11' In the case of real values of Xk, equality holds in (12) for the functions fez) = F i (C:.) Cv
I
and = I(~ omitted for ()
=
n
TI
(11)
tions,
m
I
n."
~11~lak'IXkXI'=:;; 1, n
2:
1, and
2,"', m and l
=
1, 2,"', n.)
where m
2:
Xk
{m
t1-
l~ II
k-
.
-/-, 1X '1 I
2
Let us prove two mor~ theorems for functions in the class
2
n
2: 2, on the circle
II I
1'1 = p, C.
Theorem 6. Let n denote an integer =
such that
1,
C.
--\-)2(n-l)
=
'I
(14)
where A is positive and depends only on nand £v'
except coincidental ones (v
=
v' ).
=
1,' . " n
Proof. In (6), let us take for av v' tWO systems of numbers a; v'
,
"
and
a~ v'
F(Cv)---F\~"'1
n
v'
-a., v'
c. - c.'
"i,v'=l
II ~. n
1 _ _1_
a.~. 'Y,+ct "y ,
(15)
p, we have
€,,~I===pn(n-I)
III ,€
-
c•. 1 === A1plt(It-I),
- V-:f..'
I . .*.' III
1 -- -1- === A 2 e.e-.,p2·
- - :::::e: mm
1 e.e~,p2
~l 1,
C
I
11
=
2: 2. Let
1,"', n. Let F denote a function belonging to 2. Then,
1 )2/n
II 11- ~II-,. C,
p . ..
V::F V'
(r \
1'1
=
I (I - y v=~~:nF(e.c)I~Ap
1,"" n, where
In the derivatives, v and v
n
1, v
=
2.
2
I~ (1 -
kvi
on the circle
p > 1, we have
F(C.)= F(C.,)
_
y-:f. y'
and arbitrary 'v' v
p, v = 1, .• " n, this yields inequality (14). This completes the
proof of the theorem .
and xt are arbitrary complex numbers (for k
Theorem 5. For functions F(t;;) €
I'v 1=
with
2 1 Xk 1
125
§2. SHARPENING OF DISTORTION THEOREMS
IV. EXTREMAL QUESTIONS
124
{ -
distinct numbers such that
=== I, V, v' === 1, ..., n,
I Z 1=
I
n-I
for
k v I = 1.
2. Let £v'
V =
1,"', n, denote
Then, for functions fez) €
v· === 1, ..., n.
v=l, .. n
where A is finite and depends only on nand
'1:1=- v',
S on the circle
< 1, z ) I '=:;;~'?I'" min \f(€v
for v === v·, V,
r, where r
2:
(18)
(v.
Proof. Inequality (18) is obtained from (17) if we apply the latter to the func
We then obtain the inequality n
2
2
FCC.)=FCC.,) I~ II (1- _1)2-,. II \1- ~\I-,. C, c.' _ IC,I C,C., II .*v' \ v=1 v*v' 2
(16)
tion F(D = Ilf(z), where' = liz, and to the points £v = llz v and then replace (v with '£ v • We note that the exponent 21n in the right-hand member of inequality (18) cannot be replaced with a smaller number without additional restrictions on the
IV. EXTREMAL QUESTIONS
126
§2. SHARPENING OF DISTORTION THEOREMS
function fez). To see this, consider the function
the form
z fez) =
II
_
II (I-2hZ
lD (C) = log F
)2/n
(~~
=~
(Co)
-+ log (- Co) + log ( 1 <Xl
=
h=1
In
I z \ < 1,
n
. 1
a,
+ ~ ~m( n ~ h=1
_
tk~
n k=1 ~
I•
I - 2 h z
n k=1 ~ II -
2hZ
EkZ
~
Iz I =
Consequently, as z moves around an arbitrary circle
where 0
< r < 1,
in the positive direction, the function arg fez) increases mono
r,
<Xl
CQ
T
1..
This means that fez) €
s.
p211
Now,
p )2n _ ( pi ) n;r --log 1 - Tfor°
11=1
If we now let p approach 1, this yields <Xl
If(s,r) 1= n
r
II II - £kE,r 12/n
~
!
Ar
11=1
(I ~ r)2(n
h=1
Now we have, for 1
where A is a positive number depending only on n and (v' v = 1," ., n. This proves what was said regarding the sharpness of inequality (18). =
2, we obtain from inequality (14) the following inequal
I:
I
P (~1) -
F (~2)
C1 - C 2
I::>: 1- ~ 0--
I.
This theorem, which>
was proved by the author, 1) has found numerous applications. Here, we mention the direct elementary derivation (given by Milin)2) of inequality (19) and also of inequality (10).
P (Co)
C- Co
I.:;:: ~ Ian (Co) I = ~ v;l\ an (Co) I· '1"""::1 "'-,n ~ """ I ~ In ~ 11=1
y n
11=1
ro
<Xl
~ 1/ ~ n I an (Co) 1 n=1
which, as we let
I(0 I
approach
=
..
•
!
11=1
n I ~ 1211
~ y'-IO-g-(1-1 L/2) .log (1 - m") 1(\
Ilog P (C~ CCo) I~ -
=
I
p, yields
log ( 1 -
p\)' ICI = ICol = P> 1.
(10')
we now replace the absolute value with the real part taken with a minus sign,
Specifically, the function (() regular and univalent in 1
J
ro
<Xl
P (q
(1- I C~ Ill)'
1
It
n=1
~ I ( :s;; ~ n
11=1
we have
ity for F(() €
l..
11=1
(Xl
" n I an (~o) [2
Iz I < r
a closed Jordan curve. But then the function w = fez) maps the disk
We note that, for n
ro
"., l (._ ~ )n
we have the inequality
conically by an amount 217; that is the point w = fez) describes (one time around)
< r < 1,
0
11=1
= rei¢;o
< r < 1.
+ 10g (_ C) _
~~W-oL-~";"-n2. (_p )2n) I Co I •
S -- ' I t (_ ~ n I anpIn
2
1
where z
univalently for any r in the interval 0
Zo·)
Therefore, the area S of the bounded ddmain bounded by the image C of the I P circle I z! :: p, where 1 < P < I(0 I, is ~qual to
)
=2 ~m(I+~Z)=2~ 1_1~12 >0
the domain
"" cn n=1
we have
iJargj(z)
for 0
~ an (Co)
127
=
log [F(() - F((o)), where 1(0
I> 1,
is
\ (I < I (0 \, and its expansion in that annulus has
1) Goluzin [I946e}.
2) Lebedev and Milin [1951], inequality (11.1).
we obtain inequality (19). We note that inequality (11) can be derived in the same
••
---:--:-:__ ~=--:=-_~==:·~-""",-~_~---=-_===·::;;c.=-~---==:_-_::::::--=_=--==_-_---==-~~--,".;-"",-,--"",_~",,~==,,_=_-=~~,,~=,~=.:;·=:;=c.=-::-
-~· ,-~"",=-:::-_c_-,:,::;:::-_-=-~--'--'-",::::;::",,,,,,,-~---,;;;:::~~,~~,>;;~·_~:
__:c.
§ 3.
IF has the same value for functions that differ from each IF in the subclass 1° of functions F«()of the class I such that F«() ~ 0 in 1(' > 1. That this maximum is attained f~'lows from the. normality of the class 1°. Suppose Since the quantity
Extrema and majorizations of the type of the dislortion theorems
other by a constant amount, this maximum coincides with the maximum
The method of variations also enables us to establish a numbet of results of the distortion-theorem type and to calculate completely the extremal functions. 1) Furthermore, in the investigation of extrema of certain quantities, we ate able to
tha t w = F «() E: I
ascertain their ranges of values.
Theorem 1. Let n denote a positive integer. Let Yv' v
° is one of the ~'xt.rema.lcf~ctions. Let us suppose that the
image B of the domain
We shall stop here t6 go into only one theorem in detail.
I(I
> 1 under the mapping w
=
1, " ' , n, denote
(25) of
I
have the sharp inequality
F* =
n
~ F(C.)--F(C.,») ~ ( ffi\( ~ 1,1.,log C.-C" ~-.:. 1,1.,log 1., • .,'=1
." .,'=:;;-1
Equality holds in (1) only for functions de fined in the domain
I(I
1 \
ce)'
§3
n
.,
w~
wo)
(F. woHF., C. - C.'
+ 0 (A
B
)
)
~
(F.-Wn~!F.,-Wo)
T.T.,
)+O(A
2 ).
Since Wo can always be chosen in such a way that the last sum is nonzero, we
(2)
obtain IF. > IF for small A> 0 and suitable arg A l ' But this contradicts the extremality of the function F«(). Consequently, the domain B cannot have Let us now turn to the varied function (26) of §3 of Chapter III with m
(2' )
F (t.) - F (C.')
C.-C.'
I~-
n
"\"1 - I ~ 1.1.,og " .'=1
(1 ---=-, 1)
IF. =IF- Am ( A1 " (3)
A -
1 "II.
~
~
,,'=1
(
FO)B
T.T; CoF~
-
1
F~
Fo)(F.,-f'o)
A ( 1
=
For this function we have
C:F~
n
~
FO)B
CoF~
~
T.T.'
co-c.
C~,F-., _
Co-C.'
F. - F.,
"11',,'=1
+~(C;JB
n
'11,.,'=1
~ T•T., (F. 7:I
n
Proof. Let us find the maximum of the quantity
IF=ffi( ~ 1.1., log F(CC)=[:C.,»)
1(I > 1.
•. and with (1 replaced with (0, where
C.C._ which defines the range of values for the summation on the left. Thus, for arbi trary given Yv and (v' v = 1, 2," " n, this range of values is a disk.
i
C.F~
C.'F~')
T.T.' ~--=-~~."-l
+O(A
i
),
'II, ,,'=1
(4)
(Here, all the ratios representing indeterminate forms must be replaced with the
""
for given Yv and (v' v = 1, 2,"', n, and for F«() ranging over the entire class
1
[Here and in what follows we shall write for brevity, Fv instead of F«().]
> 1, we have the inequality
n
J
(l-AA
. exterior points.
.-1
~
(F.-F.,)
'II,v'=1
n
1.1.,og I" .:. .'=1
T.T.,log
=IF-Affi(A 1
> 1 onto the entire w-plane with a cut along
.ffi (~l.log (w - F (~,» )=const.
I(I
n ~
> 1 by the
the curve defined by the equation
Furthermore, for
by using formula
n
.=1
I(I
(
" .'=1
n
These functions map the domain
ffi
v
F(C) - F(C ) ~( 1 ) ~ 1.Iog C-C.· = - .i.;,l.log .1-CC~ .=1
1°
of Chapter III with m = 1 and with wI replaced with Wo.
(1)
equation ~
F «() has an exterior
=
. point wo. Then, let us form the varied function F. «() €
n complex numbers, not all zero. Let (1' "', (n denote distinct complex num For this function, we have
bers in the domain I(I > 1. Let F denote a function belonging to I. Then, we n
129
§3. EXTREMA AND MAJORIZATIONS
IV. EXTREMAL QUESTIONS
128
ratios of deriv'atives.) If we replace the last term in the large parenthetical ex
I.
pression following ~ with its complex conjugate, we conclude on the basis of the arbitrariness of arg
A 1 and the extremality of F «(), that the quantity in the
parentheses following lR after we divide by the common factor A 1 must be equal
1) Goluzin [1947 ].
f;'
/
A l' we would have IF* > IF' which would contradict the extremality of the function F «). This leads to the
to zero because otherwise, for suitable choice of
n
ffi ( •.
condition
n
~IT.T" + "~=lT.T"
~
C+C. -C .'po ··C-~., C+~") •P' •C-C. F. -F..
~ O.
n
n
'~~F~s
131
§3. EXTREMA ANp MAJORIZATIONS
IV. EXTREMAL QUESTIONS
130
!
'II, ~/=1
1.1.' (P.-Fo) (Fv,-Fo)
+
!
qF~
n
If we denote the left-hand memb er of this inequality by 8
T;h'
", '11'=1
~ i;'(v' C6 -C.
i.o
on the circle
C~,P'., _
Co-C.'
!
T.T.'
v, v'=1
~C,,-I P -F
C;;Cv-1
. .'
Since ~ in this equation is an arbitrary point in the domain
'0
1'1
> 1,
c,~:=~(,.)).
I"
=
1, we have
A(q=B(~)~O.
It follows from this result that, if the rational function A (() has zeros on the circle
1'1 = 1,
all these zeros are of even multiplicity because the increase
in arg A (0 as , moves around a small circle with center at any of these zeros will, by virtue of the principle of symmetry, be equal to twice the increase in arg A (,) as 'moves around an arc lying in
with h > O. From this we obtain, for small h,
I"
> 1 and this last increase is
equal to an integral multiple of 2". Therefore, the function
'a
real on
0/'(00)= 1 +h.
I"
=
VA'W
1. Turning now to the differential equation (5) (with
is regular and
'0
replaced
with '), we note that twice the sum in its left-hand member can be represented in
By means of the function 0/('), let us define the following function in the class
F J (~) = F,I"(~\ (0/ (C)) = F (~) _ h (F (~).....I-~F' (~) 1J--CelG +ce~6) I
~o for different
to vanishing of the right-hand member of ( 6). Thus, on the circle
+ h) (~eia + Ce~"8) + 2h
~o.
(6)
real values of (). The quantity IFe attains its maximum on the interval 0 ~ () ~ 2" at () = O. Therefore, (dIF/d())!ezo=O, which, in expanded form is equivalent
if we
1'\ =
t=~ (q=~ - h~ ~ +~::a + o (h 2).
do this, let us again consider the function Fo = e-ieF(eieO €
1. To show this, let us consider a function t = 0/(1,) where 0/(00) = 00 and 0/' (00) > 0, that maps the domain 1'1 :> 1 univalently onto the domain I t 1> 1 with a small radial cut issuing from the point t = e- ie. This function is obviously obtained from the equation
teta + t~ia = (1
tI
M "( 1-
,
Let us show that the right-hand member of this equation is equal to zero. To
(5)
replace with " we obtain the differential equation that the extremal function F (,) must satisfy. We denote the right-hand member of this equation by A (,). Let us show that A «) ~ 0 everywhere on
1.
A(t)-B(t)=l~ (,
C.F~ C.,F~, -----
n
=
we have 8 (,) ~ 0
On the other hand, on this circle we have
F. - Fv'
", ,,'=1
the form
n
(! F.~F .=1
+ 0 (h2).
r
Coosequently, this equation can be written in the form
For this function,
IF, =IF- hffi(
1'I
(0,
i
n
ry
T.T..
! v= 1
", .,'=1 n
~ T.T.'
+ 'II •
.,'=1
C P' C+C. -C P' C+C.,) v 'C-';._;~, v'C-C.,
+ O(h
F 1.
k'
=
VA (~) .
v
If we solve it we get the following dependence of F(') on ,: 2
) ,
n
!
.
where C= e ie . Then, by virtue of the extremality of F( 0, we obtain, for all , on the circle \"
T.log(F.-F)=-
v= 1
=
1
which holds in
I" > 1.
~ y' A(q ~C =C(~),
(7)
§3. EXTREMA AND MAJORIZATIONS
IV. EXTREMAL QUESTIONS
132
-',
If we take an arbitrary boundary Wo of the domain B and denote by (0 the
corresponding point on
1(I
= 1 under the mapping
and woo In this sense, equation (7) also holds on
:a on the circle
ffi (C(C)) =
\(1
1(I
=
1. Also, since
~ (aCa~C) ) = ffi (iCC (~)) = ffi (i VA (C)) =
= 1, where
WF
(= F -I(w), we conclude by
passing to the limit that equation (7) holds (with Wo instead of F) holds between
'0
C.-C
.=1
=CC,.)
for given Yv and ~' v = 1, 2,"', n, and for F(() ranging over the entire class completely covers the circle
0,
n
".
1
_
!
!W!=-
.J.
(9)
i,i.,}Og(I- • CC ,,'=1
Let us now consider the quantity W; = F.p
(1 -cr) 1)1 II
= C (~) will have a constant real part on
1(1
~ (i' log (~, -
C) -l.Jog ( 1 -
.=1
(8)
C~
I(I > 1
I(I > 1 and hence it cannot attain its maxi > 1. From this we conclude that the left-hand side of (8) is a constant everywhere in 1(\ > 1. Then, by considering its behavior at (= 00,
IWI=-
and
WF(C),p
I
if F(()€
1
~ 1,1.,jOg(1-~)
For n
e is an arbitrary real number, we II
!
l.i"IOgF(CC~=~~C.,»)~-
'1'=1
which, by the virtue of the arbitrariness of
e,
lI.
1.1,.tog(1-
cd ..).
'1'=1
implies inequality (3). But this
also shows that the set of values of the quantity
Wl~-
!
'. ,'=1
i,l,'IOg(I-~),
=
1, inequality (3) takes the form
\ log F' (q I ~ -log (1 -1,1 12 ),
(12)
and inequality (1) takes the form, with YI = e ia
ffi (e 2i 1
onto the w-plane with cut along an arc of the curve
defined by the equation
ffi (eilL-and for F «() ranging over all the functions F«() = (+ ad( + '" € l. Obviously, this maximum coincides with the maximum
ei"log(yw-Ft which also holds on the citcle
yW-F 2 )= ~ yQ(C) ~C,
-
1(1 = 1.
of the quantity
f F= ffi (e gi " (F (C t )
From this we conclude, just as above,
among all the functions
Therefore, among the extremal functions of this last problem there are functions
yw-F2)=c+2lt,
-
without zeros in
where c = const and t is real.
image
the boundary of the form (19). This completes the proof of the theorem.
log
1"
B of the domain
following differential equation for
Theorem 3. 2 ) For functions F «() = ( + all ( + ..• €
where
Q«()
C
in
I(I
h-h-)
1 ( Km)
,c
I(I
l, we have the sharp
(m) ( 1 )' Km
2_ 2 E
-=
~1 Y (i-XS)dx(i
kllx l ) ,
E(k)=
~ -V I
1-- k x S dx . I-xs
=
If we divide (21) by
ffi
1(- (II,
(e i"
Since the function
is regular in part on
I(I
I(I
c
)? -
I(I
= 1. If we now
=
(21)
const.
C1C)
we get
C
.. / .
=cV
(22)
(C - C1 ) (l -C 1C)
/" .. IC (1" (C) - 1" (C »' V (C-C )(I-C C) 1
1
> 1 and, obviously, continuous on
1(' ?
1 and since its real
= 1 is given by formula (22), this function itself in
1(' > 1
is deter
mined by the Schwarz formula (the integral is taken in the counterclockwise direction): I" ... (CCF(C)- F(C 1 ))
1) Goluz:in [19471. 2) Theorems 3 and 4 were proved by the author (see Goluz:in [19431'11 ).
0 on
1
e
Inequality (20) defines the range of values for the quantity F«()/( for arbitrary
IF= ffi (e 2/" F (C))
(C-C J
C(1" (C) - 1" (C 1 ) \ (C - C1)(1- c1Q)
o
Proof. Let us look at the problem of maximizing the quantity
Q«() ?
1,
IC-C t I2 =
... (20' )
given ( in the domain \(\ > 1. Consequently, this range is a disk.
Q (C),
function F«(). We need only consider the case (I? 1. Then, we have, on ' "
J
=
Resting on this circumstance, we can construct an explicit expression for the
(20)
1
1"1
ffi (e i " yF (q - F (C t )) = c,
I 9
Si"
= 1 the equation
> 1. Here we us e the notation K(k)=
F «() does not have exterior
=
I(I > 1:
is a rational function such that
,~\
E
w
solve this equation, we conclude that the extremal function F«() satisfies on
use of elementary functions.
1 -2 + I - JTr
F «() in
C2F'2 /
radius nor the center of that disk can be expressed in terms of (I and (2 by the
F(C)
F «() denote one of them.
1(' > 1 under the mapping
(C 1) - p (C s) C1 - Cs
has, for arbitrary (I and (2, a disk for its range of values. But neither the
inequality
Let
points. If we then apply the varied function (26) of the same section, we get the
We mention without proof!) that the quantity
= F
I" > 1.
By applying the varied function (25) of §3 of Chapter 111, let us show that the
When we solve this equation for w, we obtain a parametric representation of
W
F «() € l. (Here a o is the constant term in the La urent €l.
expansion of F«().) But I~ has equal values for nonconstant functions F«()
that all boundary points of the domain B satisfy the equation
ei"log(yw-Ft
ao))
-
e
V
_
(C - C1 )(1 - C1 C) -
-
c 21ti
\
~
IC'I~I
-. (
V
c'
-C'+C dC'
(C' - C1 ) (1- C1C') C' - C"
+ let·
(23)
----~-----
13B
'~--~~~-~~""""""--~~-, ~.~---~-~;;~~:-- ~~'l'-;;,;_i:.;;~":""·'·1111~"l111j~!~!d=~J!'~~..:o".",",!!!!~,J'&~'!!!'1!"'" -~_,_ ,-,-::±~"~~;!,,.,,~-'02?;,!,~
IV. EXTREMAL QUESTIONS
§3. EXTREMA I;NO MAJORIZATIONS
From this we obtain FC(). To determine the real constants c and c 1 and also
If we make the substitution-x.-=1Cl(' in these two integrals and use the
notation (20') for complete elliptic integrals, we obtain
F((I) - a o, we use the normalization of FC(). Specifically, if we understand the radical appearing in (23) as referring to that branch that approaches 1 as ( ' --> ""
and equate the constant terms and coefficients of II ( in the expansions
F(Ct) -
cxo=Ct
of the two sides of (23) about (= "", we have ei~
~
c
--i
- - 21t
dC'
y~' (~ I _~') (1_1''-1'1") + I Ct, I t' 1=1
eia. ( 1 2i~ Ct+c;+CXO-F(Cl»)=-~ ~ 'I=l 1t
IC
00
"/""1'"
"'t\.11
"',..,\.
1/ (I
C') (1 - CtC') that branch that is positive on the right-hand side of the cut (0, 1/(1). Since the integrands in (24) and (25) are regular in the cut planes, we can deform the paths
t}
+lct,
,~
eia.
I
1
2i~ ~ Ct + ~ +
CXo - F
\
1
F (Ct) - CXo = Cl
+y + t
. 2 (e-ua. -
J yc'(Ct-C')(r----:.c1C') 1) -.,.°1------ t
.~
dC'
~
yC' (C 1-C')(1-C IO
> I,we have the sharp inequality
K(_I)
(1- E(I~I)\)'
41CII '"=ICI2-11\,
I CI
K(_I)'
(29)
I CI
Tnequality (29) detennines the range o[values o[the function CF"(C)IF'(C) for arbitrary given ( in '(I > 1. Consequently, this range is a disk. Proof. When the function F C'::) belongs to the class 1\'0
~'dC'
ICj2-1
(27)
F tr)=
I
c;
ICj2-1
F(C)
where the radica Is are understood to mean their positive values. By equating the.
1'1
~ in
~F"(C)+4ICII-2_ 41CI3 E(rh) , ;: :
~:
real (and imaginary), parts of the two sides of (26), we determine c and c l ' We then obtain from (27) an expression for FC(I) - a o :
Theorem 4. For functions F (Q €
(26)
1
C1 ) 2c \ C'dC' (C t) =;- ~ JI C' (C _ C') (I _ ~IC')-' I
(28)
range of values of F C()I( is proved just as in Theorem 1. This completes the proof of the theorem.
1
d~'
(1- :~~~\)
for (> 1. By virtue of the arbitrariness of a in (28), we obtain inequality (20).
As a result, we obtain
c \
~ and arbi
That this inequality also holds for complex ( in the domain I(I > 1 follows from examination of the function e-ieF(e ie (. The remaining conclusion regarding the
1/(1 to 0 along the upper side of the cut and then from 0 to II (I along the lower side of the cut (0, 1/(1)'
~=~ ~ YC'(C t -C')(1-C IC')
'{"'(F~)+C-+-l, and we denote by Pp the maximum of Pp,F overall F«()€I. By vinue of the normality of the class I, the maximum Pp obviously is
the extremal functions in that problem is obviously F p «()' It follows from the
solution of the extremal problems. This method can also be applied to various
n Ia. -*.,
(3)
up for the function Fp «(), attains its maximum Pp ' For the system of these
problems
functions of the class
I. we denote
attained. Let Fp «() denote one of the extremal functions. We denote by (II,P' for v = 1, .• " n, the points on the circle I(I =: p for which the quantity (3), set
§4. Application of the method of variations to other extremal
§ 3,
(2)
n IF(C•. p)-F(C." p)1 .*.,
(31)
(+ a l / ( + ..• E I and F«(),j c in 1(1) 1, then, byapply cl € S, we obtain inequality (20) in
In all the cases considered in
n(n-l)
\
the theorem, we define the following: for a given function F«() € by Pp,F the maximum of the product
ing (31) to the function fez) =: I/[F(I!z) -
I(I > 1.
.*.'
1(1 > 1.
IPlr\_n \;Pln_n
Proof. In addition to the quantities PF and P mentioned in the statement of
the inequality
+ + 1-lz\-2 KE (I(\zl) z I) l~2 (I-K(lz\) E (I z I) ) .
141
attained by some function w = F «() that maps the domain 1(/ > 1 onto the entire w-plane with analytic cuts and that satisfies the differential equation
C2 [F'(l:)}g.!
+... E S
at the point (=: - z, we obtain in
j(z)
(30)
then, by applying (30) to the function
. I (C+Z) I + zC -/(z)
dF VARIATIONS
taken over all pairs of disti1leLnumlf~rs 11, 11' taken from 1,"', n.) Let P denote the least upper bound of the quantity PF over all functions F «() € I. Then P is
+ 2 ~ 4-21z I" + _4_E(1 z i) I
= z + C2z2 + •. ; € S,
Second, if fez)
§4. METHOD
C., pF~, p - 2v=l=v' ! (Fv, p-F."p)(CC., p-l)
(1)
(4)
where FII,p = FP«lI, ). The right~hand member of this equation can be simplified further. Note that, if we replace one of the values (11, p with (11, peia. in the expression (3) written
1) See Goluzin [1946£, 1947, 195Id]. 2) Goluzin [1947, 1949c 1.
for Fp «() and for the values (11, p' where
11
= 1, ... , n,
we obtain a function of
142
IV. EXTREMAL QUESTIONS
a defined in the interval
-TT
< a< TT
143
§4. METH9D OF VARIATIONS
that attains its maximum at a= O. Therefore,
we would have Pp -
0, which is impossible since it is easy to find a positive
the derivative of the logarithm of this function vanishes at a= 0 and this leads
lower bound for Pp that does not depend on p (calculating, for example, Pp,F'
to the condition
where F «() == ().
m(t~' C.,P~~.P ..:;.. F., p F.,. P
)=0,
If, for the sample made, we take the limit in equation (5), written for F
Fp ('), as p -
I"
.'=1
where the summation is over all v'
= 1, .' •. , n excluding v' = v. Since this
> 1: [~F' (~)P
equation holds for all v = 1,"', n, we conclude that the numbers A.
!
F
C., pF~ 'II,
-F."p P ' p
1
I
It is now easy to show that
':::F" takes the form ,1, _ ,~,
..,
.:/=.,
=n(n-l)-2C
1'1
~ .*.'
(IC., p IS - I ) C., pF~ p -' (F.,p-F.,. p) (C-C.,p) (C'C.,p-l)
2) 'v,p' v
=
F«() €~;
1,"', n, converge to some 'v' v
4)
1'1 > 1, under the mapping
1, p'
> 1,
F«() we have, from some point on PPloF - £ ~ PPloFpl 2£ ~ Ppl ,Fp' - 2£ >= Pp ' - 2f. Consequently, P> Ppl - 2£. On the other hand, for these same values of p', we have Pp I ~ P. This shows that Pp I --> P; that is, such that Fpl (,)
-->
=
F«() maps the domain
1'1> 1
onto the entire
w-plane with analytic cuts. The analyticity of the arcs constituting the boundary of the image follows from equation (2). We denote this image by B. That B does
w = F('));
converge to certain finite numrers. This last is
we have (1-
,dF
,,~
It remains to show that w
We note that the points
a ' for v = 1,' . " n, are all distinct because, otherwise, for the chosen p v
not have any exterior points follows from the fact, that, if it did, we could, with out changing PF , add on to B a two-dimensional segment and then for the func
possible, because, for example, from the second of inequalities (15) of §3 for
F(') € ~
.=1 0 for v = 1, ... , n: We write equation (6) in inte
the function F (,) exhibited above is extremal.
3) Fv,p' v = 1,"', n, converge to a v (points on the boundary of the image of the domain
V.*., ! ,..
greater than 1, we have P
p-l: to a function
=
do this, we note that, for given
Thus, we can choose a sequence of functions that has the following properties as
1'1> 1
v
extremal function in the problem mentioned in the formulation of the theorem. To
can be applied to the functions Fp (!;,)' treated as functions of the p,arameter p> 1.
p
C
(6)
is impossible. Thus equation (7) reduces to the form (1).
(5)
for w = Fp in > 1. Now, by virtue of the normality of the class ~, the condensation principle
1) F (') converge in
~ (C~·C.)2' c.=const.
gral form:
~
Then, it can easily be combined with the preceding term. As a result, equation (4)
«,)
n
=n(n-l)+C
~ C.,pF~, p ..:;.. (F.,p-F",p)(CC.,p-l) •
!
1_ .•.
n,
v= , ... ,
=
F(c:> in
! ,.. ,., , '=F"
.'=1 are all real. Therefore, the last term on the right side of equation (4) is equal to
[~F' (C)]\1
1, we obtain the following differential equation for
1
tion w
=
F* ('), with IF' (08) I < 1, mapping
we would have
IF: ("") I < 1,
I" > 1
onto the modified domain B*
by Lindelof's principle. But then the function w
>=
\
\ \
F* (~")/F: (De) €
ber of (8) is equal to 12V3, which exceeds 16.
2 would give PF a higher value than the function F ( 1
extremal function has an expansion
under the mapping
F(q=~+IXo +
F('),
I(al -
a~) (al - ~s) (all - as) I ~ 12 V3,
with equality holding for the function
(9)
l1Jj=1.
Q
q= -
(8) over the points aI, a2, and a3 mentioned in the theorem and all functions at the problem considered in Theorem 1 with n =
3. One of
the functions providing this extremum must, in accordance with Theorem I, satisfy the differential equation (2), which for n
~2[F'(qlll._._,
=
~~~
all)(al -
as)(a~ -
The fraction on the left side of (0) cannot be reduced further. To see this, (a2 + a3)/2
= al'
Then, since la2 - a31 S 4, we would have lal - a2 \ S 2, lal - a3 \ S 2, and, consequendy, P
S 16,
(12)
Now, let us find a bound for
for the left-hand member of (8).
= aI, that is, that
as) I~ ~ 14 f.a y + 3~· 4IX~ I ~ 4( 4S 1IXll s + 3~ I ~ I~)·
Q = 43 \ IXI IS
Here, a I ' a2, and a3 are boundary points providing the corresponding maximum
suppose, for example, that (al + a2 + a3)/3
6~.
so that we have
(10)
.,=1.
=-
of the discriminant of the equation x 3 + px + q = 0, which is equal to 4p 3 + 27q2,
I (al -
+
a1 + aE as ,,_,., ~,~,.,
ala~aS
- 4IX1>
With regard to the left-hand member of (8), its square is the absolute value
3 can be written in the form
F (C)
(11)
is a polynomial with zeros x= aI, a2' a3' Then,
p = ala2 + alaS + a2 aS=
Proof. Let us consider the extremum of the left-hand member of inequality
F(,) € 2. We arrive
... ,
_ _ a1aS+a1aa+aE aa _ a1aEaa al4 ,txO.
Let us use this function to define a function in the class S:
1(.
U.
We note that a function w = lj;(z) that satisfies the conditions lj;(0) = 0 and radial cut of length equation
19h ( fo
2 (4)8 VIv {zo~f' (e)} + eioh (A)8 VIv{ !oe f'_(e)} ) + 0(1 hI2). ~ ~ e-~v ~ ~ 1-~v v=1
~l- ei0i. { z~: ;~
011
v
v- e
{Z;r'
sists of a finite number of analytic arcs and that equation (19) holds for w the circle I z I = 1.
Cv
n
L f(e)2}
v
It follows from this differential equation that the boundary of the domain B con
8
n ro (
o
v v
and consequently,
+ 1
}
= V~l (e 191 C + ei l
i,l;
1 - 1
z
v=1
with m = 1
(4)2 {zoe!' (e)} + h( fo, \8 { zoe f' (e)} + 0(1 h 12) zofo ezofo I 1- zoe Zo
f (e)8
n
We have
-{ f' (e)2} h fee) _ f(zo) v -
{
° with
~ 'Iv /(e)-f(z) v
which is impossible. Thus B has no exterior points. Let us now consider the varied function (24) of
(18)
o
'=1
0
Since the expression in parentheses is a polynomial in
{f. (E)}. = c v
v
n
(h 2),
and, consequently,
l/w3- 1
o
(-4-)1 V IV{ !oe1f' (e) } ) + 0 (I h 12 )= 0. zofo ~ 1-zoe v=1
~"
For small
" (-4)1 V Iv {ZoE!' m} zofo ~ e-z v v=1
___ n
=
v-I
of
)1
f(o/ (z»
f* (z) = lI(O) =f(z)
+ hf (z) -
+
, 1 zei'P hzf (z) 1 _ ,~I 0 and they exhaust all extremal functions in the sense of the characteristic of the images B.
Let us now apply the method of variations to the investigation of a particular
Suppose that fez)
extremal problem for the class Sa of functions
f(z) =CtZ
+ c~z~ +..., Iz I < 1
that are regular and univalent in the disk finite values a 1> ., • , am ,.where m
~
C
1. Let us look at the problem of maximizing
Theorem 4. A function w = fez)
C1 z + C2 z2 +... € Sa for which ICn I, where n is
=
a given integer ~ 1, attains its maximum maps the disk
I z 1< 1
onto the entire
w-plane with cuts along a finite number of analytic arcs and satisfies in the disk
n
> O.
R (f(z»
P (f(z»
z2f' (Z)2 = n
+ ~ (n _ ~
'I)
(C n _. z-'
• = I
Cn
+c~_.Cn z'),
(25)
~k=1 c k zk is one of the extremal functions such that
~
§ 3 of Chapter III,
N~~e can
easily obtain a differential equation for the function fez). Since
§ 3 of
the function\ (29) of distinct
zl,j···,
Chapter III is a function in the class Sa for arbitrary
ting M = 1/ in that formula
{f* (~)}n = Cn
I z 1< 1
in the disk
Zm
I
+ h f (~)
m-3
R (w) =
P(w)= Il(w-ak)'
~
and sufficiently small A, we have, set
til
II (( (e) -
ak) )
m
k;1
II
-
(((e)-f(Zk»
k=1
h
!
A~(Zk), {ej' (e) }
k= IZk! (Zk)
11
m
e-
\Zk
11
_
+ Ii ~~ Akf(Zk) {eSf' (e) } + 0 (\ h I~) Zkf'(Zk)2 1- Zke
where m
when we replace w with f(z)
and take WI = W2 = .•• = w n = 0, we can show, following the same reasoning as in the proof of Theorem 3, that the domain B does not have exterior points.
the differential equation n--I
=
Let B denote the' corresponding image. By using the varied function
obtained from equation (27) of
and that do not assume m given
for given n ~ 1 with respect to functions of the class Sa (see Goluzin [1946f]).
Iz I < I
,,~,:
f,?,,;/
theorem.
Ic n I
-
k bkw ,
n
k=1
let> (01 n denotes the coefficient of z n in the expansion of the function et>(z) about z=O. But since ~({(~)ln)~cn, we have the b k bidng constants. In equation (25), the right-hand member is real and non negative on the entire cir cle
Iz i =
where
k=-n-I
k=1
1, and the boundary cuts referred to are with
a suitable parametric representation w
=
wet) (t real), the integral curves of the
differential equation R(w) pew)
(dW)' + 1 =0. dt
(26)
m
I
II
rn(, h f (~) I m
(f(e)-ak) )
II
m
~k = I c k zk € Sa, we have the well-known inequality \c ~ I ~ en for the function f(z)/ c 1 = I. k= I C~ z k
I c n I has
and consequently,
1z 1/(1 + I z 1)2
a finite upper bound bec au se , if f(z)
I Cn I ~
en
I c 11.
(((e)- f(Zk» n
I c 11 < «1 + r)2/r) I a I.
Iz I =
r at which If(z)\
It follows from this that
I c II
fer
(e) } (Z)I C
k .. - Zk n
m
+Ii~ Ak/(Zk) {e2f'~e)} )+O(lh\~)=O. 1- Zke n
.'- zkf'(Zk)1
k=1
=
If we apply the familiar inequality If(z)/d ~
to a point z on the circle
~ z. P k=1 RJ
k=1
Proof. The existence of extremal functions is obvious. Specifically, the quantity
h ~ Akf (Z1l)
_
k=
< Iall,
we obtain
~ 4 I all as r
-->
1. Conse
quently, I Cn I:os:;; 4en I all· By virtue of the normality of the class Sa' this upper bound is attained.
If fez) is an extremal function, the function fee eLi z) for arbitrary real a, which has the same image B, will also be an extremal function. Therefore, there
If we replace the next to the last term with its conjugate (which leaves the real part unchanged), we obtain m
I
II
rn (h [ f (~) k = I
(f (e) - ak) )
m
II
m
_
(f(e)--f(Zk» n
~ Ak!(Zk) ({ ef' (e) }
.'- zkf k= I
e-
(Zk)1
Zk n
k=1
_ {e1-f' Zke ~e)} )]) + 0(1 h I~) = n 9
Because of the arbitrariness of arg h, this leads to the condition
O.
-
154
IV. EXTREMAL QUESTIONS m
(
I
II (f(E) -ak)
(~) k= 1
1
_
m
II
(f(E)-/(Zk» n
m
'\' Ak! (Zk)
I' 1- zkE
(fEr (E)} _ { E ~E)} ) = O. E- Zk n
(27)
n
= znc~
R (/ (j) Z2£' (Z)2 P (f ~;z» J
1
~ zkl (Zk)1 k= I
1
+... + nCn + (n -
where
m-3
k=1
Since the first term on the left does not vanish identically for all values of •• "
we can choose.
Z m,
let us change
Z m'
Z 2> •• "
Z b •• "
Z 1,
Zm
At =
Q (f(z»
considering it a variable z. Then, the A k take
'- ~__..
Bk .1:/_
(k=2, ... , m),
\
II (W-/(Zk))'
ak)
II
(f(e) - I(Zk»
1 n
~ (IEI'E- (E) }
P (f (z» Q (f(z» zl' (Z)ll
n
I
II
(f (E) -
I (Zk»
1= !
k=-n
"\' +k=2 ~
Ck .1:/_\
.I:
_
/_
\
-
Bkf(z)k,
V
~
so that we finally obtain
o/(z) where 0/(0)
k=2
=
0 and
wi
+ ze i9 + o (h
1
hz,
_nM
h
(nc n + 2 ~
2
).
+ ze_"Ini6 I' (z) + 0 (h it (z)
=
)
~k=1 c~l) zk, we have
kCke(n-kli6)
k=1
2
+ 0 (h
2 ).
From the extremality of the function fez) we have )R(c~l) ~ c n
I (z) - I (Zk)
,
which,
because of the smallness of h, leads to the condition
!
Dkl(zl,
k=O
~ [Z~~l + ~~~1
1
n-l
C n
.
= - . ~ ...
= -
1
obviously belongs to the class Sa' By setting
C"
=
I < 1 with a e- i e. For this function, we have the
0,
n-l
m-2
Ell (E)} _ {E 2 I' ~)}
{ E-z n 1- zE Il
=
It (z)=/(IjJ(z))=/(z) -
m
k=1
we have
but also on the circle
Iz I < 1
C~) =
-1
n
0, that maps the disk
small radial cut issuing from the point w
The function
Ell/' (E)})
1
I
=/ 1 except at a finite number of points. Consequently, the boundary of the
Z n
- { 1- zE
II
Z 1
expression in
m
f(~) ~ =
z
simpler form. To do this, let us consider a function w
k=1
m
1
The differential equation for the boundary of the domain B can be put in a
k=2
form
I
conclude, therefore, from the analytic theory of differential equa
domain B consists of a finite number of analytic arcs.
and the Bk are constants. Therefore, condition (27) can now be written in the
IT (f(E) -
a~~ b Ie' We
tions,,!that fez) is regular not only in the disk
m
Q(w)=
k= 1
~-
Admittedly, this differential equation contains the unknown constants
/
m
11 (W-ak)'
Thus, the function fez) satisfies the differential equation (28) in the disk
Izl:e 1. I
(I(~)
(28)
k=-n-l
Ck .1:/_\
where
P(w)=
+...+ CtZn-t,
such that this term is nonzero. Now, fixing
Ak =
,
1) Cn_tZ
k bkw .
~
R (w) =
Z b .••
the form p (f (z»
155
§4. MElHOD OF VARIATIONS
nC n
+ 2ffi ( ~
kCkei(n -k)
6) ~ O.
k=1
If we now set e iB = z, we obtain an inequality expressing the nonnegativeness of
the right-hand member of equation (28). Thus, on the circle
+ ... + nc n + (n-l) Cn_tZ +... + CtZn-t] ,
I< (f(z» P (f (z»
Iz I =
1,
z2f' (Z)2 :::>- O. ::0
It follows that, with a suitable parametric representation w = w (t), where t is
156
IV. EXTREMAL QUESTIONS
real, of the boundary of the domain B, we have for this boundary the differential equation
R (w)
pew)
where
" ~tl
k=l
ie
Let us stop for the important special case of n bound for
I C 1 I = I[ , (0) I.
=
bec~~se ['(0)
i~
This quantity represents the radius of the disk onto
= cp(w) normalized by the conditions cp(O)
= 0 and cp '(0)
= 1.
It is called
above concerning the bounds for the coefficients with n = 1 thus apply to the 00
and a given
finite system of points al"", am' The corresponding extremal function belong
1
where R (w)
=
I
;;01bk wk.
P(j(z»
If z
=
(Z»)I =
J(z)
rp (w)
This function is distinct from [(z)
IC I I
!c n I for
does not necessarily increase when the domain B is
/I
1 ,
over all possible functions
is the inverse function, we obtain the
h (z),
TI I/k (D) I k=l
for k
=
1, 2; ... , n?
In the case n = 2, the answer to this question was given by Lavrent 'ev [1934] (on the basis of his variational method). It consisted in the sharp ine quality If~ (O)/~ (0) I~ I a1- a~ \2,
R1 (w) rr -w!P (w)
= ...
in B. Consequently,
in which equality holds only for the functions [1 (z) and [2(Z) that map the disk
In (jl (W)=
r , / wabo(w-a +b w+ '" +b (w-a!) ... (w-a
J Y
m_1w m-
1
1)
1
m)
dw+ bm•
(29) '~l("
1z I < 1 onto the halfplanes about the common boundary of which the points a 1 and a2 are symmetric.
This yields an explicit analytical expression for the function cp(w). The con stants B 0, B 1, •. "
C
is also easy to show that Lindelof's principle does not apply to
J=
differential·equation· 'P'.. (W). If (!p)
1 n I.
[/(0) ~ (0).
ing to the class Sa satisfies the differential equation obtained from (25) with n = 1:
Rdj(z» (zl'
> 1, the function
In conclusion, let us examine the following problem on disjoint domains: Let aI, a2 .' .• , an, where n ~ 2, denote n distinct finite points in the w-plane. Functions w = h (z), k = 1, 2, ... , n, that are regular in the disk I z I < 1 map that disk univalently onto disjoint domains B k including the points ak, for k = 1, 2, .. " n, respectively in such a way that h (0) = ak for k = 1, 2, ... ,n, The question arises: can we speak of the maximum of the product
the conformal radius of the domain B (d. Chapter II, §2). The results obtained extremum of the conformal radius for domains B not containing
= e- 21T (n-l)ilm
n > l/, that is, that broadened.
1; that is, let us find a
which the domain B containing the point z = 0 is mapped univalently by the func tion z
for n
> 1 and nil (mod m). If fez) is one of the normalized extremal func
~I>0 a normalized extremal function for
n = I C n I e- . Let us apply the above reasoning to [( e i e/ n z). If we replace z with e -iel n z in the formulas obtained, we obtain the assertion of the theorem (!c n I is included in the coefficients bk ). C
C
C
tions, then the function [I(z) = [(ze21Tilm) e-21Tilm belongs to the class Sa and is
~-_l
Consider the function fez) with
I nI
> o.
n > 0 does not always have a unique extremum. Let us show this with an example. Suppose that ak = ae21Tkilm (k = 1, 2,,·, , m), where n
bkwk.
R(w)= ~
157
uniqueness of the extremal function fez) normalized by the condition (0)
fez) normalized by the condition
m-3
TI (W-ak),
§4. METHOD OF VARIATIONS
We note that, in the case of the extremum of
(dW)! dt + 1 =0,
m
P(w)=
,
Bm remain undetermined.
A formula of the form (29) was proved earlier by Lavrent'ev [1934] by starting with the principles of Lindelof and Montel. l ) These principles lead to the
The existence of extremal functions is easily proved. Specifically, since the function gk (z)
=
(h (z)
. ~ot assume in the disk
- ak )/[~(O),
Iz I < 1
belongs to S for k
the values (ak 1
follows from Koebe's theorem that (ak 1
41
- ak)
a k 1 - akl . This proves that the quantity
1, 2, ... , n, and does
(0), where k 1 ~ k, it that is,
If; (0)\ s
J has a bound independent of the form
h (z) for k = 1, 2, ... ,n. From this and the normality of the family of functions h (z) that we are considering, we conclude that the least
of the functions 1) Grotzsch [1930] studied the extremum of the conformal radius for the class Sa on the basis of other considerations. He also pointed out the analytic form of the extremal function in a certain particular case.
- ak)/ [~
[k (0) ~~;
=
upper bound J0 of the quantity J is attained.
.",=:€~=-""
----------
---~~-
,-
-
.-~.
,
-"'------"-----------
o
,
158
---
~=-----=~~,,~~~~~~~-;;;:,"==~~::~:~'!!::::::::~~_:::~:::~~~::-=--"---~~~~~
=
!k (z),
for k
=
;'
1, 2,"', n, denote a system of extremal func
LindelOf's principle ( p. 30), the domains Bk are such that the union of their
:8: (Ik (Zk) :8: (Ik (zk) _
zlk (z)
closures U k=lBk coincides with the entire w-plane.
-
•
Zk (I;' (Zk W
h
Let us prove certain variational formulas that we shall need. Let n denote an integer greater than 1. For fixed points ak, where k
+Ii
II
II (w-a.) .--:...=_11....:.1
2
z , 1k _.,(z)
~
~
""
1 ~ko~n,
_
II (w-w.) '1= k o
where h is a complex number and the w lJ ' for v
1,2,···, n, v.j k o are distinct points. This function can be represented in the form =
W*=W+h(W+AO+ ~
w:...·w.\,.
From this we easily see that, for sufficiently small
I
I h I,
the function w* is uni
for v = 1, 2,' .. , n, v.j k o are removed. Therefore, if w lJ = f)z lJ)' for v
disk
+ h :8:
(lk (z) _
=
I. (z.))
(31)
and f'ic(O)
= ak,
k
=
1, 2, " ' i n.
h (z)
=
constructed in accordance with formula (3), are also
II
J*= Dill:' (O)I=Jo 1 ~ko~n,
for k
11 +m fh (~----:'-.!;----"-k~_(ak_--_a_.)_
-
(30)
II (ak -I.(z.»
k=1
.=1
'1=k II
is, for k = k o, regular and univalent in the disk
§ 3 of
+ 0 (I h \2),
(Ik (Zk) - I. (z.))
easily obtain the equation
'1= k o
=
1_
1 -zk z
I h I onto disjoint domains
h o (z)
.=1
(30) has a simple pole z
)
admissible functions of the extremal problem that we are considering. Now, we
(Ik (z) - a.)
.=1
Ik (z)
for small
with the function
II
n (z) =
Iz I < 1
then the function
II
(zk) -a.)
1, 2, ... , n; k.j k o , let us now construct the varied functions (31) with zn ~ 0 for v = 1, 2,"', n, v.j k o , which, together
valent on the entire plane from which sufficiently small neighborhoods of all points
lJ
(lk
which is regular and univalent in the disk I z I < 1 for small values of I h \: also f'k (0) = ak' The functions w = f'k(z), k = 1, 2, .. " n, thus defined now map the
From the functions
)
.*ko
I z I < I,
:8: ( :8: .=1
.-1
Equations (30) and (31) are the desired variational formulas.
.=1
1, 2, ..• , n, where
I. (z.))
__
Z-Zk
• =tk, k o
.=1
W lJ'
.
'1=k, ko
1, 2,"', n, consider
=
a.)
I
.=1
the function
+h
159
§4. METHOD OF V ARIA TIONS
tions and let Bk denote the corresponding extremal domains. On the basis of
w* = 'w
...
'-""'-~~ _o~.".,.._...,.,""':::~c':."..~:=_::::===-::=
'",;,
~;_
IV. EXTREMAL QUESTIONS
Now, let w
~~.;~"''''--
--~;:;~~_=_:c
- --------- -
--------_.:--------
--,
I z I < 1.
+
For k.j k o , the function zk in the disk I z I < 1. Consequently, the theorem in
Chapter III can be applied to it. As a result, we obtain the function
:8: (lk (z) - a.) n (z) = Ik (z) + h :8: (lk (z) _ I. (z.)) .=1
·4:.ko
IT
.=1 (lk (Zk) -
(Zk»2
k=1
k=tko
n.
a'))J --
'~I (Ill (ZII) - I. (z.))
+ O(lhI2)} .
.*11, ko By virtue of the arbitrariness of arg h and the extremality of the functions h (z) for k = 1, 2, .. " n, this last relation can hold only if the expression in the large square brackets is equal to zero, that is only if II
II
II
.=1
L z: (11
L :*~ k=
I
II
.=1
"fko
(ak~a.)
(ak
-t. (z.))
II
II
II
+~
1
zZ (Ik(z k»'"
k=
I k =/=k
0
(lk (Zk) -
a.)
=0.
.-1
IT
.=1
'1=k, ko
(lk (Zk) _ I. (z.))
(32)
160
IV. EXTREMAL QUESTIONS
§4. METIlOD OF VARIATIONS
Consequently, this condition must be satisfied for an arbitrary choice of the points Zk,
for k
=
It follows from (32) that the circle
I I = 1. Z
I Z I < 1 and for arbitrary k o = 1, 2"", n. functions !k (z) are piecewise-analytic on the
1, 2,"', n in the disk
To see this, let us fix all the pointsz k> k = 1, 2, .. " n; k ~ k o ,
with the exception of one of them
Z
kI = Z
tion (32) becomes a differential equation for z~ (fk J (Z»i
< n).
(where 1.:s k I
for k = 1, 2, 3; that is, the functions w
!k (z)
for k
=
3
II
(ak- w)
11=1
zsw,a
We see that condi
_
(aI-as) (aj-aa)
-
a1 -w
j) (aa-a 8) + (aa-aaaW
(aI-as) +c + (al-al) al-w '
(z», 1
Of cdurse, the constant C may depend on ai, a2 and a3' However, we are not
theory of differential equations. Let us look at some particular cases. The case n
2 can be carried through
=
able !~o determine its value.
t. 0
and C = O.
to its completion without difficulty and we shall not stop to do this.
Let us consider separately the two cases C
Now let us look at the case n = 3. In this case, equation (32) with 'k o = 1 yields (a 1 - as) (a 1 - as) (as - a1 ) (as - as) (a 1 - Is (zs» (a 1 - la (z,» (as - Is (zs» (a 2 - la (za»
In the first case, equation (35) reduces to the form
l"Ptw) w'
+
3
II
3
(as - a1) (aa-as) (aa -Is (Z2» (aa - la (za»
+
1 I. (zs) -/a (zs)
If we multiply through by the difference [2 (Z2) -
(fs (zs) - a.)
z:
3
=
.=II
from ai, a2, and a3' For extremal domains B k , the union U~=IBk coincides
(fa (Z8) - a.)
with the entire w-plane. Therefore, the point w = De lies on the boundary of cer
I
zl (f~ (Za))lJ
_
0
tain of these domains Bk' let us say, for example, on the boundary of B I ' Let us
-.
denote the point on the circle
and decompose the first
A(/a (za»,
fdz) by z
as) (a 1 - aa) a1 - I (z)
+ (as -
=
1 corresponding to it under the mapping w =
I~
~ /1
2
z
=
z., we have
(A 2 - J3 are not mapped by every (z) € S OntO convex domains. Therefore 1)
This means that the disks function w
+
(z) =
Sr
thing,
I-Iz
1- 4z
1z I
re i¢, everywhere on the circle
I
we have
that are mapped by an
possessing the property that arg w always increases
we note that the disk
S r is indeed mapped onto a convex domain if r 2 - 4r + 1 ~ 0, that is, if r S 2 -.;3. Since this is true for an arbitrary function w = fez) € S, we see that Re ~ 2 -.;3. On the other hand, for the function f (z) = z/(I - z)2 € S, It follows that
Sr
fez) of the class S onto domains that are starlike about
'~omains
on that circle. But inequality (6) of §4, Chapter II yields, when we replace the
1-41
1z I
as a point w move~ around the boundary of such a domain in the positive direction.
where z
(Zl" (Z») + 1 ~ 0 f' (z) ~
(ZI" (Z») + 1 ~ !' (z)
1.78···.
a~
absolute value on the left with the real part taken with minus sign
ffi
=
The limit of starlikeness. We define the limit of starlikeness for the class S
if and only if
:tp (; + cp + argf' (z)) ~ 0 on the circle
167
§5. LIMITS OF CONVEXITY AND STARLIKENESS
z1
Sr
r exceeds
Rs
properties are possessed by mappings of disks of the form functions w
=
fez) that belong to the class S when
I
by means of ' 2)
To do this,
we introduce the concept of generalized starlikeness. We shall say that a closed domain is of the type D n (a) if every point of it can be connected with a point a by a broken line that lies entirely inside it and that consists of no more than n rectangular segments. Thus, a domain of the type D1(a) is a domain that is starlike about a point a. 1) Grunsky [1934]. 2) Goluzin [19351.
._-
----
--
...
--
_.
_;
-_._-_._---------.
-
.....
-_...........
---:
·1 ... · ··.'. ~J
Now, let us d~note by Dn (where n
=
I, 2",,) the least upper bound of all
p such that an arbitrary function of the class 5 maps the disk
Di
,;:;;
Iz \ < I
onto spiral-shaped domains. Furthermore, D1
Rs
=
Di = tanh 4' The exact values of D" for
a..
possible lower bounds for
n>
(3)
is mapped by an arbitrary [unction w
=
_
I+EC
g () ~ -
lel
D", we see that
2D,,/(1 + D~) ~ D 2 " and hence that
2D"
'+D."
2m (Ez)( 1- d~';:;; d~ -
e(I - d~)) IE 1 (I - d~) z I _ d~ I HI +, ... ,,,.
=
exists an a of moduius unity such that the image of the disk (7) includes
this is true for arbitrary dn
onto a domain of the type Drr([(~)).
2
then the disks (7)cover
n rectilinear segments. We conclude from this that 2d,,/(l + d ~) :::: D 2 ". Since
...
Ir\~d '0 -.;: '"
eI'" -;~d"'ll ; ; : I. - fz I'"'
-
lal = I,
in it and that consists of no more than 2n rectilinear segments because there
I-DI ",;:;;
Iz
2m
I and
point [( ad,,) by broken lines lying entirely in it and consisting of not more than
In~
'+ I+D~ 2D."
(1-_D,,)2
I+D" ,
which proves the first of inequalities (6). To prove the second of inequalities (6), we set ~ = ar in (4), where
d n < r< I, and n >0. Since I EI (I - d~) -t- d" (I -I Ell) _ I EI ± d" 7 1- d~ I Ell 1- d~ i E12 - 1"'-_--I--d-=-,,--'I--=E ,
From (5), we get in succession
Izl -
Ia \ =
both these points, with the result that it is possible to connect them with the
Consequently, this function maps the closed disk
Z=
'(
in (4), where
+ d~). An arbitrary point in the image of this
disk can be connected with the point [(0)
w = [(z) maps the closed disk
9.
~ = ad"
n ~ O. We then see that every function w = [(z) €
< D", into a region of the type Dn (0). Therefore, the function
2
(6)
for arbitrary' m ~ I and n ~ 1.
completely the disk
[(z) € S into a domain o[ the type D,,([(~)).
I( C+3 ) -/(;)
e>/(I -
'-DI"+m~(I-D,,),,,. I-Dm I + Ds,,+m ~ I +D" I +Dm
Iz - I ~d~ I,;:;; I ~d~' d,,< D"
Proof. If [(z) belongs to the class 5, so does the function
I(z -
(I - I E)1)2 IE IS)2 ,
(I - d~
. . ., .
its boundary. If we let a move around the circle
I
(n= 1, 2, ...)
that is, the disk
,
onto a domain of the form D,,([( ad n )). The disk (7) includes the point z = 0 on
I
I ~I < 1..
I-D s",,;;;;(I-D,,)2 I +D" '
\I I +DI"
I are unknown. We shall give the greatest
E(1- d~) d" (I-I EIS) Z-I_d~IEI!';:;; I _ A S I C I ! '
where dn
\
We use the following
Lemma. The closed disk
:s d",
d~
d~1
-
oR
To dlrive the first inequality, we set ~
for arbitrary ~ in
+"
"
and, consequently, we have from (2)
I(,
IE 12 (I
We shall now use this lemma to prove the inequalities
.....•
D", ,;:;; D3 ,;:;; • • •
for ar!:>itrary n because there exist functions of the class 5
that map the open disk
E[2 ,,;;;; I - d~ I EII
from which (4) follows immediately.
Obviously, from the definition itself, we have
__ n",g~~~~~:r~~~~i§~~~i~~~-:C::-iQ~7~"~~::~.:.!%:=&..:..:_~~.~~::':::'.~._~~
IV. EXTREMAL QUESTIONS
170
and is tangent to both its boundary circles. If we now let a move around the
circle
Ia I =
question is given by the theorem on p. 174. The proof of that theorem is based on
1, the disks ( 4) completely cover this annulus. Let us now choose
two lemmas, which we shall now prove.
~ in such a way that (j ~I - dn )/(1 - dn I~I ) = Dm , where m> O. This yields I~I = (Om + dm)/(1 + dmD m) < 1. Then, by the same reasoning used to derive the first of inequalities (6), we see that the image of the disk is of the type D2ntm (0) a~d consequently that This yields
(I
Lemma 1 (Grotzsch's principle).l) Suppose that we have a set of n disjoint
simplx connected domains S k, for k
I z I s (I ~I
+ dn )/(1 + d n I~I) ~I + d n )/(1 + d n I ~p .s D2ntm .
annulus-;
=0
=
curves each of ':which has a nondegenerate arc for k
= 1", " n, contained in a rectangle with sides oflengths A and B (see Figure 11). Suppose that the boundary of each of these domains is a closed
onto some rectangle in such a way that four given
-
-
W
the vertices of the rectangle. We prove this last
-
A
1'////'£;/;;;/;;1;
-..
x
accessible points on the boundary are mapped into U I
B Figure 11
assertion as follows: Let us first map our domain onto the upper halfplane and let uS denote by a I, a2, a 3, and a4 the points on the real axis into which the four given boundary points are mapped. (We can arrange for all the ak to be finite.) We then map the upper halfplane onto the rectangle by using the function
Jordan curve that contains a nondegenerate segment (that is, a segment not con sisting of a single point) of each side of length A. (The domains Sk can be regarded as strips extending from one side of length A to the other.) If we map
~
"
~
«
w=oV~-~~-~«-~~-~·
Then, the points a I, a2, a 3, and a4 are mapped into the vertices of the rectangle.
174
175
§6. COVERING OF SEGMENTS AND AREAS
IV. EXTREMAL QUESTIONS
fo (z)
As a consequence of Lemma 2, we obtain the result that the ratio of the sides of
strips Ski and Sk" whose images Lk' and Lk" under the function w
the rectangle is determined uniquely by the mapped domain and the points that are
will have a common boundary segment on the rays L k and will together constitute
mapped into the vertices.
=
Lz
Theorem. I) Let f(z) denote a member of S. Consider n rays issuing from the point w
=
0 forming equal angles each with the next. On each of these n rays
consider that point of intfrsection of the ray with the boundary C of the image of the disk
Iz I < 1
under the mapping w = fez) which is closest to the point w =
Then, at least one of these n points is at a distance no less than w
=
\Il1;;"
o.
from
o.
L1
I
Proof. Let us suppose first that the boundary C is an analytic curve and that the theorem is not valid. Then there exists a system of rays L k , k
=
1, "', n,
that issue from w
=
0 at equal angles such that the boundary points on them that
are closest to w
=
0 are all at a distance not exceeding
d < \11/4
from w
=
O.
c
The function
w=!o(z)=dl/Lf
z
2
az+ ... , a=dl/=f
< 1,
(1 +eiazn)n
for suitable a maps the disk
Iz I < 1
Figure 12
onto the plane with cuts in the form of n
halflines lying on the rays L k issuing from the points A k = e-(a+2kTT)iln for k = 1, " " n that include the point w = 0 on their extensions. Since a < 1, it follows that for r sufficiently small, the image function w
=
f (z)
S
contains in its interior the image So of a
where Q> 1, under the function w
=
I z I = r under circle I z I = Qr,
of the circle
the
fo (z). Let us now partition the annulus
Qr < I z I < 1 into a system of strips by drawing an atbitrary set of rays issuing from z
=
0 in such a way that the following three conditions are satisfied:
1)
The set includes all rays lk containing segments that the function w = fo(z) maps into segments OAk and all rays that are bisectors of the angles between adjacent lk (see Figure 12);
2) it includes ;111 rays that the function w
maps into curves tangent to C;
=
fo (z)
a strip L k ' U L k" which extends from So and returns to So. The entire w-plane exclusive of So and, consequently, the image of the annulus r < the function w
I) Lavrent'ev and Sepelev [1930, 1937] a
< 1, under
=0
0' k" &
where R
~()~tained in B that is starlike about w = 0, that is, that has the property that a
f(z) to the parts of the boundaries s k '
and Sic" contained in Qr
I z I < R,
lq
by a rotation through an angle
2"k/n (see Figure 14). Let us denote by p" (k = 1, 2"", nm) the greatest
~
B kIt = log
Jr .
distance between w
=
0 and points of the part gk ,?f the star lying between lk
and l"+1 and outside s (for k
= 1, 2,'" ,
nm, gnm+1
= g1)'
By means of the func
tion ~ = log (wi p), we transform this system of domains gle into a system of do
Consequently,
ak' 21' - +011") -~--
~( b
k,
bll "
log _1_ Qr
mains h k lying respectively in rectangles with sides of length log (Pk / p) and ak . lK"
When we compare this with (3), we obtain 2'lt' :>2l' . Q. ~ 1. --1-::0-I ' I.e. loglog r Qr But this contradicts the fact that Q > 1. This proves the theorem for the case in
lit
which C is an analytic curve. In the general case, we need only apply this result to the function f(pz)/ P €
S, where p < I, and then take the limit as p -- 1. By the same method we can Figure 14
also prove the following theorem on the covering of the segments.
1) Goluzin [1936a]. A generalization of this theorem was obtained by Bermant [1938, 1944 ].
----_._. -~--~~_....~". ._-""'~=---~--~--"""""'--...,--"~. ---~-~---~~---_."----~~-~------§6. COVERING OF SEGMENTS AND AREAS
179
IV. EXTREMAL QUESTIONS
178
Here, ~~:. ak = 217 where amk+ 1= al. On the horizontal sides -of each of these
n
1:
~-
rectangles is a boundary segment (see Figure 15).
n
Ck~ -~
1
k=1
11
n
IT
Ck'
Ck~O,
(5)
k=1
we obtain .S:~+iry
I
n-I
V
'k=O
\,
Qt/(
~vn-I 'II_log
~
n
log Pmk+l -P-
\
Pmk+l -P-
n-I
!
k=O
n!
log Pmk+1 _P_
k=O
\
\
\
Consequently, (4) yields
ooi
n-I
' .. §
n log R ~ max V log r
lDg.ftJ p
we have, in accordance with Lemma 2,
< (Rp)n r Therefore, by taking the limit as r
ak ~ --.!!:!:.
bk
P
n-1
When the domain h k is mapped onto a rectangle with sides ak and h such that the boundary segments mentioned above are mapped into the sides of length ,
Pmk+1
'k=O
or
Figure 15
bk
1
max 1
II
Pmk+I'
k,;""O
0, we obtain for some II in the closed
-->
interval 1:S [1 :S m
log Pk
n-I
P
Rn
-: ~
R =- ' 1=1 I og~ r
'Y
2"
___ :>-: min
R =I ogr
1
~ 7.
I P
p(
n-
1)
I-f. m+ 1
n
. ,., ,
'-i
1=1
If we now use twice the familiar inequality
CT. 1
n-I
~ Ak~ ~ I
n
~ - I = -2". mm 'n 1 olog Pmk+1 k=
P
(4)
P/l - E, PI2 - E,' " ,
Consequently,
k=1
m
.... I Pmk+1 k=Oog---
.. "
.
A2" ", An and hence are respectively greater than
P
Therefore, since ~T=1 al = 2TT/n, we have
n-I
~1'
n-I CT. I 1.- I Pmk+l k=O o g -
Let us now take an arbitrary system of n rays issuing from w = 0 at equal angles (2TT/n) with each other and intersecting respectively gIl' grn+/l"" " ' , g(n-l)m+ 11 • The parts of these rays lying inside the star are of lengths
Pmk+l.
-ns~n(R-e).
k=O
For R sufficiently close to 1 and
f
sufficiently small, this is arbitrarily close
to n. This completes the proof of Theorem 2. As a special case (with n = 2) of this theorem, we obtain the result that there exists an interval of length arbitrarily close point w
=
to
2 that passes through the
0 and that lies entirely in the image of the disk
1
z I < 1 under the
mapping w = fez) €
that is, by setting z = rei 0 (it obviously holds also with A < 0) we get the ine quality
~ R Ad ~ ~ RAd. C
(4) ~
C'
In particular, if C I is the circle \ w I = R 0, then (4) yields
~
"!"-t I-"!" .! f'(z)=zP fez) p f;(zp).
RAd ~ 21tR~.
Let us denote the left-hand side of ( 6) by I, replacing above expression. We obtain
(5)
l=r
C
If we apply inequality (5) to the case in which C is Cr and R0 = M(r) + (, where ( > 0, and then let ( approach 0, we obtain from (2) the inequality
b
l~r
~ If(reiO)IAdB~21tAMr)A,
2~ ~1j(reiDJI' -IF (reiD) IdB·
'\ -_. Mlr Z I~ n=l
Thus, in the disk I z 1 < 1 we have
Mlr
l) Goluzin [194Sa1. Bazilevic [1951] and, independently, Milin (see Lebedev and Milin [1951}) have given a sharp bound for in rhe class S. The bound given in rhe text was found by Milin (see Lebedev and Milin (19491). For an improvement in the bounds for the coefficients of odd functions, see Alenicyn [19511. (In 195~, Gong Sheng (Kung Sheng = Kung Sun) showed that, for the coefficients of odd functions, ICn I < 2.55 (see §2 of the
lenl
I
supplement>. )
2) Schaeffer and sfencer [19431 and also Goluzin [1949c].
3) Goluzin [1948b .
f.(z)I~.oM2~0, with the same M2 • We set 7J v
I
= e
21TV
I
i/iJ., J/' =
(3)
1,"', /l. For fixed z in the
disk
1zl < 1, let us look at the values of If(7Jv' z)j, v'
=
that the greatest of these absolute values is obtained for v Theorem 7 of §2 ( with n
I
=
2) applied to If(7J v z)1 and If(7J v 1 we obtain the Il - 1 inequalities
v'
inequalities
1,"', Il· Suppose
=
II (7j" z) I ~ 1Mrr '
I
z)\' where v' -!
Vb
(4)
= 7j,fp., 1 (z) + 7j~11'-, 2(z) + ... + 7j~1p.. p.'
= 2 and a = 0, this theorem is a di
rect sharpening of Theorem 2 of §7 in the sense that it establishes the relative
,
1
Ilc nt ll-lc n llE;cn 4 Iogn, n=2, 3, ... ,
(5)
VI'
Now, let v. denote any of the numbers 1,"', Il distinct from v. Consider
(9)
where c is an absolute constant.
where v =
1,"', Il and v' ~Vl:
Proof. Here, we shall need a more definite form of sharpening of the dis tortion theorem than that provided by Theorem 7 of §2 (with n = 2). Specifically, ,
1,'7j~' =
{ 1 for s=V*, s =F v*, 0 for
"1:-'1
let fez) denote a member of 5 and suppose that the maximum value of If(z)1 on s~1
- , ... ,
(Jo,
(6)
s =F v.
the circle
I z I = r,
,
where 0
ing inequality (16) of
According to the lemma, this system has solutions. Let us denote these solutions I
n do not even approach zero for
odd n. This completes the proof of the theorem.
have
the system of Il - 1 linear equations with the Il - 1 unknowns Y v I
by Yv' . For ea c h v
Ilk +
Theorem 2.1) For functions fez) = I.~ =1 C n zn belonging to the class 5, we
+ 7j~1p..2(Z) + ... + 7j~'/p., p.(z) I E; 1Mrr , '/' =F
!
C
=
class 5.
Therefore, inequalities (4) can be written in the form
\ 7j.-!p., 1 (z)
are satisfied for arbitrary even Il and v for all n
boundedness of the even and odd coefficients of functions in the class 5. We now give bounds on the relative growth of successive coefficients of functions in the
It follows from (1) that
(7j,.z)
Ie n I ~ c n a.
We note that, in the simplest case of Il
=F vI'
193
THE RELATIVE GROWTH OF COEFFICIENTS
v with k = 0, 1, " ' , whereas the coefficients
Then, from
V I .
where M depends only on-Il'
I
§s.
.~
IV. EXTREMAL QUESTIONS
192
§ 2 to the
is attained at the point z l ' Then, by apply~
function F «() =
I
(7)
M'r IIp., '. (z) IE; (1- ~)1+"
(8) =
I z -- Zl II I
Thus, the assertion of the theorem is proved when a ~ O. That this assertion is
IE; 1_2r rll .
(10)
Let us now prove inequality (9). From
1,," , Il but
But now that we have proved inequality (3), it follows immediately on the . • baSIS of Lemma 3 of §6 that 1c n I ~ c ' n a., where c I IS the c I 0 f the theorem.
2
(z)
This is the sharpened form of the distortion theorem that we shall need. 00
v. ,ft v. But in accordance with (2), such an inequality holds for v. = v. In view of this, we obtain from (1) inequality (3) with M2 depending only on Il' a, and c.
v, we have e -II ~ 11 (r) ~ I '\(r)l. From this we obtain
(1_ ,)4
,2
tJo' ('t) - A' ('t) ~ 2e--' (I-'- ('t) -I A('t) I).
this yields inequality
(15), proving the theorem.
(4)
Integrating equation (4) from 0 to "", we obtain in accordance with (3)
The inequalities given by Theorems 2 and 2 I are undoubtedly not sharp, but they are the best that have been obtained.
00
00
•
00
~ I A('t) I d't ~ ~ I-'- ('t)d't = ~ e-' d't + ~ e--~ d't o 0 0 •
= ('I +
1) e-',
§9. Sharp bounds on the coefficients As we stated above, in addition to a sharp bound for the second coefficient, we have a sharp bound for the third coefficient for functions in the class S. To illustrate the nontrivial nature of solving certain problems on the extrema of the coefficients, we now present a more general result. Its derivation is based on
Consequently we obtain inequality (2). From what was said above, we easily see that equality holds in (2) only for the function '\(r) = ±Il(r). This completes the proof of the lemma.
Theorem 1. 1) For functions 00
f(z) = ~ cn z ll
Lemma 1. Let '\(r) denote an arbitrary real function that is continuous for
0< r < "" except possibly at a finite number of points at which there are discon tinuities of the firs t kind. Suppose that I'\(r) I :5 e -r for r ~ O. Then, by setting
11_1
belonging to the class S and an arbitrary real number a in the interval 0 ~ a we have the sharp inequality
I ca -
00
~ Ai('t)d't=(~+ ;)e-i .,
'1>0,
(1)
we have
I~A('t)d'tIE:;;(~+])e-·.
(2)
o
Inequality (2) is s harp for each given v with equality holding only for the func tion '\(r) = ±Il(r), where Il(r) = e- II for 0 < r< v and Il(r) = e- r for r> v. [We note that a v satisfying condition (1) always exists because the left hand member does not exceed e- 2rdr = Yz, and the right-hand member decreases
JO'
from
Yz
In particular Ic 3-'
2
IXC: I ~ 2e- 1- II
+ 1.
(5)
:5 3.
Proof. It will be sufficient for us to consider a function fez) € 00
to 0 as v increases from 0 to
"".J
co
00
ca-lXc~=4(1-1X)(~ e-~k('t)d'tY -2~ e-i~ki('t)d't. (6) o 0 Since the function e i f3 f(e- i f3 z ) for arbitrary real f3 belongs to S' whenever fez) does, the maximum value of IC3 - a d I in the class S coinc ides with the maximum
ad). = e i8 (r),
00
v
00
~
tJoi ('t) d't =
~
ffi (ca-elC;) = 4 (I-IX) [( ~ e-~ cos a ('t) d't
00
e-i ' d't +
~ e-'~ d't =
('I +
~) e-"
o
00 00
= } A' ('t) d't.
(3)
Now, for an arbitrary function '\(r) of the type in question, we have for r ~ 0
§ I).
For such functions, formulas (23) and (24) of §2 of Chapter III are valid. On the
Now, we have from (6), by setting k(r)
given v is easily verified. Specifically,
S' (see
basis of these formulas, we have
of the quantity R(c3 -
Proof. That the function 11 (r) belongs to the type of functions ,\ (r) with
< I,
Y
00
- (~ e-~ sin a('t) d't )'] - 4 ~ e-i~ cosia ('t)d't + 1 ~ 1) Fekete and Szego [1933]. See also Bazilevi~ [1936].
~
__
~
~
.
_
=
~
~
o
~
_
___
~
-------, 198
IV. EXTREMAL QUESTIONS ex>
ex>
~ 4 (I - Il) (~ e-< cos 6 ('t) dt )2 o
-
4 ~ e-~< cos~ 6 (t) dt
fp (z) = ~ c~ + IZ"P+1
ex>
+ I,
where p > 1, we have (he sharp inequality
(7)
I
the form (1), we have from (2)
\
m(ca - IlC~) ~ 4 (I -
Il) (\I
r
= 4 (l -
+ I)g eII)
\I~ -
I-
• -
(I -
I
+ ~ )t,-g· + I 2(1) + -} - II1 e-'J' + 1.
4 ("
(8)
\I
In particular, Ic~2)1:s e- 2/3 + 2- 1
=
ex>
+ I'
= 4 (I - Il) (~ A(t) dt)g - 4 ~ A~ (t) dt o 0 where ,\ (r)
199
§9. SHARP BOUNDS ON THE COEFFICIENTS
Ca - p 2p
C~ I ~ 2e
-2 P - 1 p+ 1
+ 1.
This together with (14) yields (13). For certain special subclasses of functions of the class S, sharp bounds for
00
~ e-' sin 6(t)dt=O', o
and that A(r)
=
(10)
e -r cos O(r) coincide, in accordance with the lemma, with the
function fl (r) (with v
k('t)=
=
{
al( 1 - a», that is, that
1"
2 e-('-')+IVI -e(.-.)
for 't~\I, for
(11)
t>".
the coefficients c n have been obtained. For this, we need Lemma 2. Suppose that a function fez) = 1+ alz + a2 z2 + ... is regular in the disk z I < 1 and satisfies in that disk the condition 'R(f(z» ~ O. Then, Ian l:s 2 for arbitrary n = 1, 2,···. J
Proof. The condition 'iR(f(z» ~ 0 is equiv;'llent to the inequality 11+ f(z)1 ~
11 - f(z)J.
It follows that the function
It is easy to show that we can satisfy condition (10) by choosing the sign ± properly in (11). Specifically, if we take the plus sign in (11) for 0 and the minus sign for
f3 < r < v,
then, by virtue of ( 11), condition (10) can be
.
written in the form ~
~e-'V I-e o For
f3
=
g(.-.)
g is regular in the disk
dt-~e- O.
at ( zj" (z) ) I' (z)
in
I z I < 1.
(24)
By virtue of the maximum principle of harmonic functions, equality
cannot hold here. It is easy to see that the converse is also true, that is, if a function fez)
= z
+
C2 z2
+ .;. is regular in
(24) in that disk, it maps the disk I z Analogously, if a function f (z) = z +
I
0'
point of the function f(z) analogously defined at z 0' On the other hand, if the limit is 0, then fez) has a pole at z =
Zo'
In this case, we take f(zo) =
both cases, the function (= fez) maps a neighborhood of the point z
00.
In
° onto a
domain including the point (0 = f(zo)' Now, if fez) is not univalent in the domain obtained from B by adjoining to it the point z 0' there exists a point z 1 € B such that fez 1) = (0' A neighborhood of the point z 1 is mapped by the function
205
§ 1.
V. UNIyALENT MAPPING OF MUL TIPL Y CONNECTED DOMAINS
206
(= f(z) onto a domain including the point (0' Consequently, to every point of a
Let us now show that the domain B" can be mapped univalently onto a circular
< 1(I < h.
"I > 1
two points of the domain B. However, this contradicts the univalence of f(z) in
annulus 1
B. Thus, our remark is proved. It shows that, when we are investigating questions
t = log z " onto the halfplane ~Ht)
of univalent conformal mapping, isolated boundary points can be adjoined to the
corresponding to each point z" in the domain Iz
domain being mapped witllOut causing any complication in the problem.
many points of the form t + 2k TT i, for k = 0,
Let us now begin with the simplest case of the problem posed, namely, the
207
domain B corresponding in the domain B" to sequences that converge to z ~ .
sufficiently small neighborhood of the point (0 there correspond no fewer than
.~
MAPPING A DOUBLY CONNECTED DOMAIN ONTO AN ANNULUS
plane ~(t)
> O.
We first map the domain !Z
> O.
by means of the function
This mapping is not one-to-one because
"I > 1
± 1, ± 2,
there correspond infinitely
The domain B" is thus mapped into a vertical curvilinear strip
case of doubly connected domains. Let us show that every doubly connected
in the t-plane bounded by the imaginary axis and an infinite analytic curve con
domain can be mapped univalently onto some circular annulus, whose boundary
tained in the halfplane ~ (t)
circles may degenerate to points.
the property that, if a point t E B l' then all points t + 2k TT i, for k
Let B denote a doubly connected domain in the z-plane. H it has an isolated boundary point z 0' then, by adjoining it to the domain B, we obtain a simply connected domain that we can then map univalently either onto the disk or onto the (-plane with the point (= that the point z = z
° is mapped into
mapped either onto the annulus 0
00
1(1 < 1
excluded. We can do this in such a way
(= O. The domain B is thus univalently
< Izi < 1
or onto the annulus 0
< Izi
R,
=
z + R 2e 2i e/ z, which
e to the real axis.
= z + a o + at! z + ... is univalent in > R and the entire boundary of
~ 21z1 in the domain Izi
Izi
,=
F(z) is contained in the
then the fun ction
~=Fl (z) =_I-F{zoz)-~=z+~+ ... Zo Zo ZoZ
above, we can give a proof based (like, the proof of Riemann's theorem) on the solution of an extremal problem: Let B denote a doubly connected domain that is contained in the domain Iz I > 1 and that has as one of its boundary continua the
is univalent in Izi
circle Iz I
entire boundary of the image of the domain Izi
,= f(z)
z + R2e ia/ z. Therefore,
with equality holding, obviously, only tor the function F (z)
B is mapped onto a subdomain of
Out of the family ~ of all functions
=
ffi (e-~i6 ell) ~ R
that annulus, to which (3) is applicable with inequality holding. Then, the ratio
=1.
Chapter II as applied to the
2,
The assertion of the theorem follows because, if we map the extended domain
B' onto the annulus
§ 4 of
~ R2 with equality holding only when F(z)
(3)
¢ «() = c ." where
with equality holding, obviously, only for
la.1
that
211
R2.
Proof. It follows from the area theorem of
Con
-R: > ,R.'
MAPPING ONTO A PLANE
that map the domain
> 1.
Consequently, in accordance with §4 of Chapter iI, the
> 1 under this function is contained
B univalently onto domains of the same type in such a way that the circle Iz I = 1
in the disk
is mapped onto itself, find the one for which the quantity
jz I > R. Then, if we shift from \z I > R to lz I = R, we obtain the second part of
zEB
has the greatest value. The function representing the solution of this extremal
!Iii:
Theorem 1.1) Every domain B in the z-plane can be mapped univalently onto
of given inclination
e to the real axis.
point a of the domain B is mapped into
with parallel rectilinear cuts
Also, this mapping is such that a given
,=
()Q
and the expansion of the mapping
function about z = a is of the form
Let us now investigate univalent conformal mapping of arbitrary multiply
_1_ +Clt(z-a)+ ... or z-a according as a is finite or infinite.
connected domains onto canonical domains of various kinds. The simplest such domain is a plane with parallel rectilinear cuts. In this case, our investigation will be based on the solution of certain extremal problems.
Iz I > R, the quantity llHe -2i e a l ), where
~ 2\zl, where
complement of the domain B / with respect to the plane is a straight line segment
§2. Univalent mapping of a multiply connected domain onto a plane
z + a./z + •••
aol
a domain B' in the extended '-plane such that an arbitrary continuum in the
problem maps the domain B univalently onto a circular annulus.
=
~ 2. In particular, IF .(1)\ ~ 2; that is, If(z) -
the lemma.
M (/)= sup If(z) I
Lemma 1. Out of all functions F(z)
1R onto the plane with rectilinear cut that makes an angle
e with the real axis,
1) Beginning with the exuemal property of [he mapping function, which was indicated in the proof given below, this theorem was proved by de Posse! [1931] and Grorzsch [I932].
and only by that Junction. For it,
!
_Ii
_.
212
~;;;;:,~h~_
..-:
.,,'"
.::::".~~,-,
..:",-..:.,._.
,;;
."",,,.
213
§2. MAPPING ONTO A PLANE
V. UNIVALENT MAPPING OF MUL TIPL Y CONNECTED DOMAINS
the existence of a solution of the extremal problem posed is proved.
which reduces the problem to the original case.
Let us now show that the extremal function fo(z) provides the mapping referred
In the case of a simply connected domain B, the theorem is easily proved.
Specifically, if the boundary of the domain B has more than a single boundary
to in the theorem. Let us suppose that the complement of the domain B with
point, then it follows from Riemann's theorem that it can be mapped univalently
respect to its image B' under the function
onto an arbitrary domain constituting the entire (-plane with cut at inclination to the real axis in such a way that z sion of the function
(= "
=
(z) about z
where a-I> O. Then, the function
,=
00
is mapped into z' = 00 and the expan
=
00
e
than a straight line segment of inclination
,=
axis and that it has the expansion Let us show that R(e- 2ie f3l) > o.
Let us turn to the general case. Consider the family ~ of all functions f(z) =
00.
t'filllil'i:
I'*
f
':,
Let us first prove that this problem has a solution. Suppose that the entire the family
3J! are univalent in the domain
Iz I > R.
with Lemma 1, for these func tions we have numbers Bt(e- 2iea l ) are uniformly bounded. of these numbers. H this upper bound is not functions fn(z) € 3J! for which the sequence
Then all functions of
'~
'( ~
1"
SHe- 2i ea l rs R 2; that is, the Let A denote the least upper bound
attained, there exists a sequence of IR(e-2ieal)l converges to A. But,
! I,I if;;'
s 2 in Iz I > R.
'!Ill'
,t,
al
Izi =
10 accordance with Vitali's
f o(z),
which, in accordance with Theorem 2 of
in B o and hence in B. Therefore, fo(z) €
§ 1 of Chapter
'(t) inverse
=
the w-plane with rectilinear cut of inclination tion
,= '(t) - =t + y/t + .•. maps
e with the real axis, and the func
It I > R' onto a different domain.
Yo
< ffi (e-
ii9
(~I
+
11»,
Now, let us look at the function w
=
B univalently and has the expansion z + of z =
w (f o(z )). This function maps the domain (a~O) + (31)/z + ... in a neighborhood
00.
Consequently, w(f o(z)) €
!Jl and we have =
at (e-ii9 II~O») + at (e-ii9 ~I) > A, f o(z)
meets all of
important extremal property of the mapping function, namely, the fact that it maximizes the quantity ~He-2ieal)' 10 the case of finitely connected domains, this theorem can be formulated as
1
follows:
theorem, the sequence of these functions converges inside B 0 to a regular func tion
,=
the requirements of the theorem. At the same time, we have established an
the functions fnk(z) are uniformly bounded in the interior of the domain B o 00.
=
00.
B I can be mapped uni
which, however, is impossible. This shows that the function
R', where
R ' > R. By virtue of the univalence of fnk (z) in B, the same inequality holds also at points of the domain B that belong to the disk Izi < R '.Consequently, obtained from B by removing the point z =
w
,=
t + Yo + y/ t + • . . . The function w('(t)) - Yo = t + (f31 + YI)/t + ••• maps the domain ItI > R' univalently,onto
at (e- ii9 (II~O) + ~1»
that converges uniformly in the interior of the domain Iz I > R to a regular func tion fo{z)/z = 1 + O)/z2 + ... such that !He-2iealO») = A. But in accordance for functions fnk(z) on
in a ne ighborhood of
that is, R(e- 2ie f3l) > O.
fn(z)/z, that is, that the sequence Ifn(z)/zl contains a subsequence Ifnk(z)/z\
s 2R'
+.. ,
valendy onto the domain It I > R' in such a way that the function
at (e-ii9 11)
This shows that the condensation principle can be applied to the functions
with Lemma 2, we have Ifnk(z)\
=
Therefore, on the basis of Lemma 1 we have
Consequently, in accordance
from Lemma 2, for func tions of the family D we have If (z)/ z I
f3 / ,
to the mapping function has the form '(t)
pose the extremal problem: Out of all the functions of the family ~, find the one that maximizes the quantity R(e- 2ie a l ).
Iz I < R.
'+
e
In accordance with Riemann's theorem, the domain
a/ z + .•.
An example of such a function is f(z) == z. Let us
boundary of the domain B is contained in the disk
with the real axis. Let B I denote
the '-plane contains the point 00. Suppose that the function w w(O maps B I univalently onto the w-plane with rectilinear cut of inclination with the real
mapping of the domain B.
in a neighborhood of z
o(z) contains a continuum other
whichever of the simply connected domains complementary to this continuum in
is of the form a-I z + a o + a/ z + ... , «((z) - a o)/ a-I provides the required
that map the domain B univalendy and have the expansion f(z) = z +
,= f e
Theorem 1 '(Hilhert). Every n-connected domain B 'in the z-plane can be
I, is univalent
mapped univalently onto the '-plane with n parallel finite cuts of inclination
D,
which contradicts our assumption that the least upper bound A is not attained by the numbers R(e- 2i eal)' Thus,
with the real axis in such a way that a given point z
I
I
.1:
=
a is then mapped into
e
-----'~.-
214
(=
. ----.--
.._.
.
._.
..
,_...
._~-_"=-==~~~=~-.,_,=.~_.==~_==-,.,==_=======~--
_-.....",.~"'=~_'__..,___..=-~-=,7"_.".,.,.","~_'""
.. .,.. "=.-.-"",,. ,"',=
~--'-_'"".""'--"-,"'.='""
;-" ,",,'C- ._.
§:z.
V. UNIVAL ENT MAPPING OF MUL TIPL Y CONNECTED DOMAINS
and the expansion
00,
I
z_ a
0
(Xl
(z - a)
+...
z+~+ z
or
'-""':';;;';: .~..,. ~""~_"""'''_'~'"''"'-"'"~-'='''''''''''''''""''''~'''''',''''"_'' _. '__ '~"''''''",,_'''",,'''''.'_'","
215
MAPPING ONTO A PLANE
Theor em 2. There exists only one function performing the mapping mentioned
f the mapping function about z = a has the form
+
.==.~~"''"'-'-~-,
m Theorem 1 '.
... ,
Proof. Let us suppose first that B is a bounded domain with boundary con
according as a is finite or not. Some of the cuts referred to may consist of single
sisting. of closed analytic Jordan curves K l '
points.
are two functions (' =
f 1(z)
and ("
=
f 2(z)
.••
,K". Let us suppose that th~re
that map the domain B univalently
onto the plane with parallel rectilinear cuts of inclination () and normalized as
Let us now look at the .question of the correspondence of boundaries under
B
univalent conformal mapping of multiply connected domains. We shall confine our
indicated in Theorem 1 '. Then both these functions are regular in
selves here to finitely connected domains that have only accessible boundary
the point z = a. On K v , v = 1, ... ,n, they assume values that lie respectively . stralg . h tInes I" C\« -i8;-') C\« on certalO -0 e '::. = C v, an d -0 e-i8;-") '::. = c v/I , f or v= 1,"', n;
points. We may assume that the point
00
is in the interior of the domain. Let B
except at
denote such a domain and let K 1, K 2 , ••• , K" denote its boundary continua. Let us map the domain B univalently onto a domain bounded by closed analytic
that is, on K v we have ~(e-i8fl(z)) = c~ and~He-i8f/z)) = c:, where c~ and c ~, v = 1, ••. ,n, are constants. It follows from this that the difference (=
Jordan curves. This can be done as follows: we map the simply connected domain
F(z)
including z =
00
= fl(z) - f 2(z) is regular in B and assumes on K v (for v = 1,"" n) values
lying respectively on certain straight lines d V' Consequently, if we take an
and bounded by the continuum K 1 onto the interior of a circle.
arbitrary point (0 that does not lie on any of the straight lines d v, we conclude
Under such a mapping, the domain B is mapped into a domain B' bounded by a
that arg [F (z) - (0] does not chan~e as we move around any of the curves K v .
circle K{ and the continua K~, ... , K~ . Then, we map the simply connected
Therefore, on the basis of Cauchy's familiar theorem on the zeros of analytic
domain containing the domain B' and bounded by the continuum K~ onto the interior of a circle. Under this mapping, the domain B I is mapped into a domain
functions, it follows that the function F(z)- (0 has no zeros in B, that is, that
B" bounded by a circle K;, an analytic curve K; and the continua K;, ... , K:'
F(z) does not assume the value (0 in B. Thus, all values that F(z) assumes in
Continuing in this way, after n steps we arrive at an n-connected domain
B lie on the straight lines d v' However, this is possible only when F(z) == const
B(")
= 0, it follows that F(z)
whose boundary consists of n closed analytic Jordan curves. The univalent map
in B. Since F(a)
ping of the domain B onto the domain B(") is a composite of successive mappings
theorem is proved in this case.
of univalent domains. Since these mappings define a one-to-one correspondence
assume that it has no isolated boundary points. If there were two distinct func tions performing the mapping described in Theorem 1', we could, by mapping the
aries defined by the mapping of B onto B(n) is also one-to-one. Furthermore,
§ 3 of
°
domain B univalently onto the bounded domain B with boundary consisting of
Chapter II, we_can show that the function inverse to the mapping
function is continuous in B(") except at the point
00.
If the domain B is bounded
only by closed Jordan curves, we can show in addition that the mapping of
closed analytic Jordan curves, find two distinct functions mapping the domain B
B
§ 3 of
°
univalently as indicated in Theorem 1', which, on the basis of what was said
onto BtnI is also bicontinuous. If we now map the domain B(n) onto the plane with rectilinear cuts, then, just as in the proof of Theorem 5 of
= [Jz), so that the
Now, let B denote an arbitrary finitely connected domain. Here, we may
between the boundaries, the correspondence between the points of these bound as in
== 0, that is, f 1(z)
above, is impossible. This completes the proof of the theorem.
Chapter II,
In conclusion, we pre sent an
we can show that the mapping function is regular on the entire boundary of the
~mportant
re lationship for the functions mentioned
in Theorem l' for various values of ().
domain B("). Combining all that we have said, we conclude that, in the pre sen t
Let B denote an n-connected domain bounded by closed analytic Jordan
case, the mapping mentioned in Theorem l' is one-to-one except for boundary
curves K l '
points of the domain B.
...
,K n and let (= j e(z, a) denote a function that maps the domain
B univalently as indicated in Theorem 1 '. Let us show that the following rela
An important supplement to the existence theorem proved above is the follow
tionship holds for arbitrary ():
ing uniqueness theorem:
ii
216
h(Z, a)=eiO(cosf.ijO(Z, a)-lsinBj" (Z, a».
217
§3. MAPPING ONTO A HELICAL DOMAIN
V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS
about z
(1)
=
b has the form
2
z 1 b +(7;) +Cll (Z - b)+...
or
Z+ClO+
a; +
... ,
(1)
The difference d(z) between the two sides of this asserted equation is a function that is regular in the domain B and that vanishes at z
=
according as b is finite or infinite. In the case in which the domain B has 'a
a. Furthermore, all the
single boundary point, this is obvious, and then the arc of the logarithmic spiral
values that d(z) assumes on any of the curves K v lie on a straight line C\I (
.,J
e
-i
e':0' J'I = const.
referred to degenerates to a point. 00 the other hand, if the boundary of the domain B is a continuum, let us first map B conformally onto the domain
Reasoning as in the p~oof of Theorem 2, let us show that d(z) '" 0; that is, let us prove (1). This relation, which has been proved for a domain bounded by
Iz 'I > 1
closed analytic Jordan curves is also valid for an arbitrary finitely connected
ping functio n z' = c/J (z) has in a neighborhood of z
in such a way that the point z 1
a+(7;)+Cll(Z-b)+... . zb
domain B. For this, it. is sufficient to map the domain B onto a domain B * bounded by closed analytic Jordan curves, apply (1) to B*, and then return to the
e as
theorem. Suppose that the point z soon as we know io(z, a) and
e and
c, the equation ~ (e - i 110 g
0=
plane with cut along a logarithmic spiral and has an expansion of the form (1). This result for simply connected domains can be generalized to multiply connected domains. For this generalization, we need the analogue of the minimal
= log (, this logarithmic spiral is mapped into the straight line ~(e-iet) = c the real axis, and the ray referred to is mapped into a
property proved in
'{ ,
I
"
0, the logarithmic spiral degenerates
e=
17/2, it degenerates to a circle with center at the origin. H we hold e constant and vary c, we obtain various curves
into a ray issuing from the origin. For
constituting the family of logarithmic spirals of inclination when we speak of logarithmic spirals of inclination
e,
e.
:il·'t )
I.
i:
of Chapter W, which we formulate as a
= z + a o + all z + ... that map the domain univalently in such a way that a given finite point a and the point 00 are mapped into the poi nts 0 and 00 respectively, the quantity SHe -2i e log F '(a» Lemma. Of all functions (= F (z)
Izi > R
Iz I > R
with cut along an arc of a logarithmic spiral of inclination
these are what we mean.
e.
onto the (-plane Here log F '(z)
means the branch that approaches 0 as z - .... 00. We now have
points, it is possible to map B onto the (-plane with cut along an arc of a loga
e in
§3
is minimized by the function F o(z) that maps the disk
In all that follows,
Let us show mat, for any simply connected domain B that has boundary rithmic spiral of inclination
But this possibility
['(00) = 00 with suitably chosen c provides a mapping of the domain B onto the
This last follows, for example, from the fact that, if we shift to the plane
e=
onto the (
As a result, we see that the function (= cF(c/J(z» = fez) with f(o) = 0 and
c defines a logarithmic
property that it is intersected by an arbitrary ray issuing from the origin at an
straight line parallel to the real axis. For
>1
by the condition F (a ') = 0 provides the required mapping.
spiral in the (:-plane with asymptotic point at the origin. This spiral has the
e to
z 'I
this problem has a solution and that the extremal function (= F (z ') normalized
with cuts along circular arcs of concentric circles. For constant
e.
1
'»
logarithmic spirals and, as limiting cases, onto the plane with radial cuts and
t
a is mapped into a point z' =a'. It now
SHe-2i elog F '(a in the class I. This problem was studied in § 3 of Chapter IV (the first application of Theorem 1 with a = 17/2 - e). It was shown there that
mapping of multiply connected domains onto a plane with cuts along arcs of
with inclination
=
plane with cut along a logarithmic spiral of inclination
Univalent mapping of a multiply connecled domain onto a helical domain
e.
Cl_1Z+Clo+~-t... , z
follows from the solution of the problem on the minimum of the quantity
In an analogous manner, we shall find the answer to the question of univalent
angle
or
= b, the expansion
remains to establish the possibility of mapping the domain
iTTI2 (z, a).
§ 3.
is mapped into z' = 00. Then, the map
according as b is finite or infinite. This is possible by virtue of Riemann's
domain B. This yields the relation (1) for B. Equation (1) yields ie(z, a) for arbitrary
=b
Theorem
such a way that given points a and b of the
domain B are mapped into 0 and 00 and the expansion of the mapping function
1.1)
Every domain B in the z-plane can be mapped univalently onto
1) Grotzsch [1931].
II~
218
§ 3.
V. UNIVALENT MAPPING OF MUL TIPLY CONNECTED DOMAINS
a domain B I in the '-plane that includes the points 0 and 00 such that an arbi trary continuum of the complement of the domain B' with respect to the '-plane is an arC of a logarithmic spiral of given inclination e. This mapping maps given points a and b of the domain B into 0 and 00 and the expansion of the mapping function about z = b has the form (z - bfl + a o + a l (z - b) + ... or z + a o + a1z- 1 + ••• according as b is finite or infinite. Proof. We may assume that b = the transformation z·
=
00
MAPPING ONTO A HELICAL DOMAIN
219
> R. If lal > R, this last inequality with z = a implies that la~n)1 ~ 21al. On the other hand, if lal < R, then the point = fn(a) = 0 does not belong to the image of the domain Izi > R under the mapping fn(z); hence, it
domain Izi
'0
belongs to the disk cases,
because, if this is not the case, we can use
la~n)1 ~ 2max(lal, R)
Iz I > R.
in
1'- a~n)1 ~ 2R. =
Therefore, la~n)1 ~ 2R.
M. But then,
,=
Thus, in all
Ifn (x)/zl ~ 2 + M/R
Consequently, the condensation principle can be applied
the sequence lfn(z)/zl, so that it contains a subsequence 1fnk(z)/zl that converges
I/(z - b) to switch from B to a new domain B. and we
uniformly in Iz I > R to a regular function. The corresponding sequence lfnk (Z)l
can then prove the theorem fex this last domain, replacing a and b with I/(a, b)
converges uniformly in an arbitrary closed bounded subset of the domain Iz I > R
and
> R. But since the values of fnk(z) corresponding to points of the domain B that lie in Iz I < 2R belong to the disk
00.
,=
to a function fo(z) that is univalent in Izi
In the case of simply connected domains, the theorem was proved at the
beginning of this section.
I' -
To prove it in the case of a multiply connected domain
a~n)1 ~ 4R and hence to the disk I' I ~ 4R + M, the set of functions fn k(z) is
B, let us consider the family ~ of all functions f(z) that map B univalently in
uniformly bounded inside the domain B with the point z
such a way that a and
and that have an
the uniform convergence, mentioned above, to
An example of
the domain B, and
expansion
f (z) =
00
are mapped respectively into 0 and
z + a o + a/ z + : .• in a neighborhood of z
such a function is f(z)
=
=
00 00.
z - a. We pose the extremal problem: Out of all func
tions of the family ~, find the one that minimizes the quantity z
-->
In this
domains in the t;plane that are complementary to this continuum contains 0 and
domain, all functions f(z) E !Dl are univalent. Consequently, for these functions, the quantity
3'He- 2i B log ['(a))
00
quantity
are mapped into 0 and
3He- 2iB
log
['(a))
00.
¢ (0)
e in such a way that
2iO
logf~ (a)) _
According to Lemma 2 of
§ 2,
=
fn(z)
1'- a~n)1 ~ 2R
=
'+
f30
+
,= ,=
'(t) = t + Yo + y/t + .... Suppose that the point t = a' corresponds to the point f o(a). The function w = ¢ ('(t)) maps the
expansion of the form
A.
domain It I > R univalently onto the w-plane with cut along an arc af a loga
e
rithmic spiral of inclination in such a way that the point t w = O. From the preceding lemma, we have
it follows that the entire boundary of the image of
the domain Iz I > R under the function w
n = 1,2,"') lies in the disk
,=
valently onto a domain It I > R in such a way that the inverse function has an
1, 2"" ,
B lies in the disk I z 1 < R.
Suppose that the entire boundary of the domain
e in such a way that
In accordance with Riemann's theorem, the domain B 1 can be mapped uni
00,
m(e-
¢ (,) maps B 1 univalently onto the w-plane
m(e- 2iO log cp' (0)) < O.
for functions in ~ by A.
=
=
0 and ¢ (00) = 00 and suppose that ¢ (0 has an expansion in a neighborhood of 00. Let us show that
Let us denote the greatest lower bound of the
opposite. Then, there exists a sequence of functions fn(z) € ~, for n -->
=
f3/' + ...
Let us show that this greatest lower bound is attained. We assume the such that, as n
Suppose that the function w
with cut along an arc of a logarithmic spiral of inclination
B' univalently onto the t;plane
with cut along an arc of a logarithmic spiral of inclination
a and
00.
is bounded below by the same number as was
calculated for a function that maps the domain
,=
fo(z) contains a continuum other than an arc of a loga rithmic spiral of inclination e. Let B 1 denote whichever of the simply connected under the mapping
00.
f o(z) holds also in the interior of f o(z) thus proves to be a solution of
Let us show that the extremal function fo(z) provides the mapping indicated
Let us show that this problem has a solution. To do this let us choose a
B' contained in B that includes a and
excluded. Therefore
in the theorem. Let us suppose that the complement of the image of the domain B
00.
simply connected domain
M. The function
00
the extremal problem posed, so that this problem does have a solution.
iHe- 2iB logf'(a)),
where log ['(z) denotes the single-valued branch in the domain B that approaches
o as
f o(z) E
=
z + a~n) + ain )/ z + ... (where
=
a' is mapped into.
m(e- 2iO (log cp' (0) + log~' (a'»)) < m(e- 2iO log~' (a'».
and that Ifn(z) - a~n)1 ~ 21z1 in the
'i.
_~
~_~
_ _.
,__.--.....,.. -----..--
-,.-._.__._-_.. --e.".,
_~
_",,,
."~
_
.~...
_
.,
_
§3. MAPPING ONTO A HELICAL DOMAIN
V. UNIVALENT MAPPING OF MUL TIPL Y CONNECTED DOMAINS
220
Therefore, ~He-2ielog ¢'(O))
< O.
Let us now look at the function w
The correspondence of the boundaries under the mappings referred to in Theorem =
¢ (f 0 (z))
domain B univalently in such a way that a and
= 00
00.
1 ' is studied in the same way as in
are mapped into 0 and
domain B has only accessible boundary points, then the mapping is one-to-one
00
hand, if the domain B is bounded by closed Jordan curves, the mapping function
+ ffi (e-
2i8
Iog/, (a»
is continuous in B except at the point z = b. Finally, if the domain B is
< A,
bounded by analytic Jordan curves, the mapping function is analytic on the bound ary of the domain B.
which, however, contradicts the definition of A. This completes the proof of the
We also have a uniqueness theorem:
theorem. At the same time, we have established the following extremal property of
Theorem 2. There exists only one function carrying out the mapping mentioned
the mapping function, which includes the distortion and rotation theorems for the
in Theorem 1 '.
family of univalent functions that we are considering. 1)
Proof. Let us assume first that the domain B is finite and bounded only by
Of all univalent functions (~F (z) defined on a domain B including have an expansion of the form
F(z)=z+cx o+ CI; in a neighborhood of z =
00,
00
Kn • Let us suppose that there exist two functions (= f l(z) and (= f 2(Z) that carry out the mapping of the domain B mentioned in Theorem 1'. These functions are regular on Kv for v = 1, ... , n
that
closed analytic Jordan curves
+...,
Kl '
.3 (e- iO log/I (z»
function defined in Theorem 1
.3 (e- iO log /2 (z» = c;,
have proved can be formulated in a different way:
where c~ and c~ are constants. Th~refore,itweset (==F(z)==log(f 1(z)/f 2(z)),
Theorem 1 '.2) Every n-connected domain B in the z-plane can be mapped
then the function F (z) is regular in B and the values that it assumes on the
univalently onto the (-plane with n cuts along arcs of logarithmic spirals of mapped into 0 and
00
curves Kvfor v = 1, ... ,n, lie on a straight line segment. We then see, just as
a and b of the domain Bare
in the proof of Theorem 2 of
respectively, and the expansion of the mapping function
const. But f 1(z)/f 2(z)
about z = b has the form z
~
b+CXO+CX1(z-b)+ ... or oz+cxo
may degenerate into points.
e=
that F(z) == const, that is, f 1(z)/f2(z) ==
§ 2.
We mention one relationship regarding the mapping functions which is analogous to (1) of
and
§ 2,
1 as z - .... b. Consequently, f1(z) "" f2(z).
reasoning analogous to that used in the proof of Theorem 2 of
according as b is finite or infinite. Some of these arcs of logarithmic spirals
e= 0
-->
Now, if B is a finitely connected domain, we can prove the theorem by
+ :J + ... ,
From Theorem 1', we obtain as special cases with
=c~,
and
In the case of multiply connected domains, the existence theorem that we
e in such a way that given points
... ,
and, on each of these curves, we have
the quantity ~(e-2ielog F'(z)) with given z E B
e is minimized by the
and given real
inclination
particular, let us show that, if the
in this case also except for points on the boundary of the domain B. On the other
Consequently, it belongs to the family !Jl and
ffi (e- 2iB log~' (a» = ffi (e- 2iB log r.r/ (0»
§ 2. In
if; (z). This function maps the
respectively and it has the expansion z + Co + c / z + ... in a neighborhood of
(=
221
§ 2.
Suppose that B is an n-connected domain bounded by closed analytic Jordan
TT/2
curves
theorems on the existence of a univalent mapping of an n-connected domain onto
Kl '
... ,
Kn and that (== j e(z, a, b) is a function that maps the domain
B univalently as indicated in Theorem I'. Let us show that the following rela
the plane with radial cuts and onto the plane with cuts along arcs of concentric
tionship holds in B for arbitrary
circles. ---1) GcOtzsch [I931a], Goluzin [1937, 1947]. 2) (Koebe [1918].)
e:
log}a(z, a, b)=eIO(cos8Iog!o(z, a, b)-lsinelogj,,(z, a, b» "2
~:
(2)
V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS
222
§4. RELATIONSHIPS INVOLVING THE MAPPING FUNCTIONS
where suitable branches of the logarithms are chosen. The difference d (z) between
assume that the domain B is finite and bounded by analytic curves Kl' K 2 , ••• , Kn • Since
the two sides of (2) is a regular function in B and the values that it assumes on the curves
Kv ,
for 1/ = 1, ... ,n, all lie on some straight line ~'He-ie()
Reasoning as in the proof of Th eorem 2 of const
=
§ 2,
223
= const.
:J (10 (z, a»
= canst,
'J (lj fC (z, a» 1"
let us show that d(z):=
= canst,
on K v (1/= 1 , " ' , n) for fixed a € B, that is, since
c, where c is equal to 0 for a suitably chosen branch of d (z). This
+ const j" (z, a)=- j .. (z, a) + const, "2 2"
yields equation (2), which is now proved for a particular form of the domain B.
io (z,
To see this, let us map the domain B onto a domain B. bounded by closed analytic Jordan curves and isolated points. We apply formula (2) to B. and then return to the domain B. As a result, we obtain (2) for the domain B.
a) = jo (z, a)
we also have on the basis of (1)
Some theorems on the existence of univalent mappings of multiply connected
P(z, a)=-Q(z, a)+k.(a),
domains onto various other canonical domains will be given in §6.
(4)
on K v , where kv(a) is independent of z € K v. Analogously, since
§ 4.
Some relationships involving the mapping functions
Equation (2) of
§3
3 (log io (z, a,
enables us to find, for a given n-connected domain B a
e if we know the
function j e (z, a, b) with arbitrary
::J (llog j (z, a, b» 1t
= canst,
"2 on K v for fixed a, b € B, it follows on the basis of (2) that
functions j 0 (z, a, b) and
j TTl 2(z, a, b). But we can also establish relationships between these last func
P (z, a, b) = - Q (z, a, b) +k. (a, b),
tions and the functions j o(z, a) and j TTl /z, a) of §2 v.hich correspond to the same
(5)
on Kv, where kv(a, b) is independent of z E K v .
donain B. To put these relations in the most symmetric form, we set
i U. (z, a) -
b» = canst,
Now, let us look at the integral
P(z,
a)~
Q(z,
a)=+(j~(z, a)+ jo(z, a»
j, (z, a», ) (1)
/J
= 2~i } P (e, z) Pc (e,
a, b) de,
2
taken over the entire boundary
and, analogously,
K=
U~= 1 K v of the domain
B in the positive
direction. 10 accordance with (1) and (2), the functions F (i;, z) and P (i;, a, b) are, for
P(z, a, b)=~ (logj~(z, a, b)-logjo(z, a,
b», )
2
Q(z, a, b)=+(Jogj~(z, a, b)+logjo(Z, a, b».
arbitrary fixed z, a, b € B, regular in B. Therefore, in accordance with Cauchy's (2)
theorem, /1 n
2
h=
d
O. On the other hand, applying (4) and (5), we have
2~{
.=1 n
These relations then take the forms
dz P(z, a, b)=P(b, z)-P(a, z),
I
=
)
:zQ(z, a, b)=Q(b, .z)-Q(a, z).
(3)
(
/
On the basis of the same principles as in the preceding sections, we can
I'~
\ pee, z) dp(e, a, b) ;.
= v~12~i ). (- Q(e,
z)+k.(z»d(- Q(e, a, b)+k.(a, b»
n
='~12~{).
Q(e, z)dQ(e, a,
b)=-2~i)
Q(C, z)QC(e, a, b)dC.
:
O
;
_
~
.
!
l
m
~
=""""' _ _
L
l
l
l
d
~
n,-._
~
__
.
__Jil_C
§4-
V. UNIVALENT MAPPING OF MUL TlPLY CONNECTED DOMAINS
224
(Here, we used the single-valuedness of the function
Q«(,
curves K v') Since
Q'«(,
z) and
Q«(, a,
not know explicitly is called the method of contour integration. I) Relations (4)
a, b) have simple poles in B at the point
and (5) on the boundary of the domain play the significant role in it. We now point out a curious analytic relationship with the preceding values of
last integral we obtain
Q~ (z, a, b) -
/1= =
225
This method of deriving various relationships between functions that we do
b) on each of the
z and the points a and b respectively, by applying the residue theorem to the
Remembering that 11
RELATIONSHIPS INVOLVING THE MAPPING FUNCTIONS
the function mapping the domain B onto a circular annulus with concentric
Q(a, z)+ Q(b, z).
circular cuts. Specifically, let us denote by (= fez) a function that maps a
0, we then arrive at the second of formulas (3).
domain B not containing"" and bounded by analytic Jordan curves K v ' lJ = lJ = 1, ... ,n, onto the annulus q < \(1 < 1 with cuts along arcs of concentric
Let us examine the integral
circles (see Chapter V,
/~=2~1 ~ Q(C,
§6,
the case 8rr/2 of Theorem 7). Suppose that the curves
K 1 and K 2 are mapped onto the circles
z)PC(C, a, b)dC
1(1 =
q and
1(1 =
1. Just as before, we
calculate each of the in tegrals in an analogous fashion. On the one hand, it is obviously equal to _P; (~ a, b).
2~i ~ P (C. z,
On th~ other hand, we can show, just as above, that it is equal to P (b, z)
/3 =
P (a, z).
h = 2~1 ~ P (C, z) (log! (C»c dC,
Thus we obtain the first of formulas (3).
Q; (z,
a, b) is an analytic function in B of the arguments a and b, whereas P: (z, a, b) is an One consequence of formulas (3) is that the derivative
analytic function of the difference function of
Q(z,
a in
a and b. a) -
a), where z and z
B and that P (z, a) - P (z
I,
I
a.
a) and
P; (z,
with (6) and (7), and the fact that
If
we let z' approach z, we see that analogous conclusions hold for the derivatives
Q; (z,
(8)
where z, z 'E B, by two different methods. Keeping in mind this fact, together
belong to B, is an analytic
a) is an analytic function of
(7)
K
Conversely, we conclude from these formulas that
Q(z',
z')(log! (C»c dC,
,
log!(C) = log!(C)
"
+ const on
K.
'~~
and
I
a).
2~i ~
The last result can be derived in a different way, specifically, from the
(Iog!(C»c dC= 1,
KI
relations
2~i
\ (log!(C)cdC=-l,
~
we can prove the follow ing formulas: 2) P~ (z, a) = P~ (a, z), Q~ (z, a) =
'11
Q~ (a, z),
!(z)=!(z') ek2(z,
!
which we can prove by examining the integrals
Is =
2k }p (C,
z) Qc(C. a) dC,
h=
21 ~
P (C, z) PC(C, a) dC
(6)
the, same way as above. That the functions
P (z,
a) and
Q(z,
a) themselves are not necessarily
analytic functions of a is shown by the example when B is the domain \z I > 1.
In this case, we easily see that
P(z, a)
I z-a I-[alil-az'
Q (z, a)
I
I
-Tlil2
I-liz
z- a .
(z)
=
z') -kdz, z'),
ef k l (z) dz - f k a (z) dz
(9) (10)
J.
1) Garabedian and Schiffer [1949 2) In the process, we need to evaluate the integral
2~1 ~ Q (C. z,
Z') (log! (e», dC,
K
and, to do this. we need, by virtue of the multiple-valuedness of the function Q( C z, z '), to make a cut in B connecting the points z and z ' and then replace rhe integral over K with an integral over a closed curve y encircling rhis cut. We then constrict the curve y around the cut. Integration by parts then leads to an integral rhat can be calculated in accordance wi th rhe residue theorem.
226
,
V. UNIVALENT MAPPING OF MUL TlPL Y CONNECTED DOMAINS
Theorem (minimizalion of area). I) Out of all functions f (z) that are regular
We note that z' appears in (9) as an apparent parameter since fez) is, in
§6 of Chapter V, determined up to a constant
accordance with Theorem 7 of
227
§4. RELATIONSHIPS INVOLVING THE MAPPING FUNCTIONS
£n
B and satisfy the conditions f(a) = (), f(b) = 1, where a, bE B, the minimum
value of the integral
factor. In conclusion, we derive an integral formula involving the function P(z, a, b).
B,
For any two functions f(z) and g (z) that are regular in
if,
we define their scalar
product as
B
r
(f,
which is equal to
lI
g) = ~ ~/' (z) g (z) dx dy.
1)= ~ ~ II' (z) I~ do,
77/ P(b,
b, a), is attained only by the function P (2, b, a) 10 (z)= P (b, b, a) .
( ll)
B
Proof. If we represent an arbitrary function f(z) of the type described in the In accordance with Green's formula
~
=
regular in B and that
f (z) g(z) dz = ~ I (z) g (z) dx '- il (z) g' (z) dy
K
(-l iJ/(Z1[W - iJ/(Z1[W) dxdy = -
¢ (a)
=
2i
B
~ ~ f' (z)g (z)dxdy,
(~(z),
B
where z = x + iy. For
(f, g)
we also have the contour representation -
¢ (b)
=
O. For this function, we have in accordance
with formula (13)
K
~~
¢ (z ), we conclude that the function ¢ (z) is
theorem in the form f (z) = f 0 (z) +
1 \
if, g)=-2i "/(z)g(z)dz.
P(z, b, a»=O.
Therefore,
(/, 1> = (/0' 10) + (Cf' 1-;) + (Cf' To) + (~, ~) =
(12)
(/0'
10) + (~,
~).
K
It follows that
If we apply (12) to the case in which g(z) = P(z, a, b) and remember that
(I, 1) ~ (/0' 10),
~ l(z)P~(z, a, b)dz
with equality holding only when (¢'
K
= ~/(z)d ~ (logj:r:..(z, a, b)--logjo(z, a, b»
0, that is, only when f(z) == f o(z). This
completes the proof of the theorem.
K 2
= }/(Z)d~ (-logj-'i(Z, a, b)-logjo(z, a, b»
As our second application of formula (13), consider a sequence of functions
uf
kJ)«() such that kJ)(b) = 0 for v = 1, 2, " ' , that are regular in lJ and satisfy the orthogonality conditions
=-~/(z)dQ(z, a, b)=27ti(f(b)-/(a», K
(k., k.,) =
we obtain the de sired formula
(f(z), P(z, a, b»=1C(f(a)-/(b».
¢) =
I for
'oJ
=
'oJ'
{ o for
'oJ
=1=
'oJ'
'oJ, 'oJ'
and have the property that an arbitrary function
(13)
in
= 1, 2, ...
f (z)
with
f (b) = 0
that is regular
B can be expanded in a convergent Fourier serie s in B:
We point out two applications of this formula. The first of these is the
I) P is contained in each of the domains An and
so that
I i') \~ + 2
'l'n
But the convergence of the sequence
Ifn (z)!
\
, "
a~n) 1
I'
r .
to a function
f (z)
to a function
f (z)
implies that the sequence la~n)l converges to a finite limit. Therefore, the func tions ¢n ( [or n
the expansions
converges, with respect to z 0' to the kernel A. Suppose that the [unctions
,=
a:
W=~2(z)=H~(z»=z+2z +
W = ~a (z) •
•
•
•
= ~ (~2 (z»
•
•
•
•
•
,z •
•
I
+ 3 :~ + •
•
•
= 1,
2, " ' , in the z-plane. Suppose that the sequence IAnl
[n(Z) map the domains An univalently onto domains B n in such a way that
[n(zo)
=
'0
and [;(zo) > 0 [or n
=
1,2, .••. For the sequence l[n(z)1 to converge
uniformly inside the domain A to a univalent [unction, it is necessary and su[
•
[icient that the sequence IBnl have a kernel with respect to
These functions also map the domain A univalently into itself. Since the domain Iz I > P is contained in A, the functions l/Jn (z), for n = 2, 3, .•. , are, in partic > p and the functions l/Jn(Pz)/p= z + naJi'pV+1z V are
We may assume that z 0
ular, univalent in Izi
2,3, " ' , which can be the case only if a v = O. This con tradiction shows that l/J(z) == z, that is, that [.(z) == [ ..(z). This last fact in turn =
shows that the sequence
Un (z)l
and converge to
0 because we can always arrange for this by
= '0 =
making translations of the z- and '-planes. The proof of the first part of the theorem is now based on Theorem I.
valent in Iz I > 1. It then follows as a consequence of the area theorem that Inavlpv+ll ~ 1, n
'0
it. Then the limiting [unction [(z) maps A univalently onto B.
Proof of the necessity. If [n(z)
[(z) is univalent in A, then [; (0) the functions
converges in z o' Thus, the sequence 1[71 (n)\
[(z) uniformly in the interior of A and ['(0) f, O. Consequently, the sequence of
->
-->
converges everywhere in the d,omain A. The uniform convergence of this sequence inside the domain A now follows from Vitali's theorem. This completes the proof
I~ (0)
Fn(Z)=_( I )=Z+ ..
of the sufficiency of the conditions of the theorem and hence the theorem itself.
In Z
One can also prove other theorems on the convergence of sequences of uni which are univalent in the domains A: obtained from An by means of the mapping
valent functions on the basis of the manner in which they ace normalized. We shall
z ' = II z, converges uniformly inside the corresponding kernel A'. Therefore, on
give one of these theorems.
the basis of Theorem 1, the sequence of the images B; of the domains A~ under
Just as above, we shall first introduce ,the following concepts: Let (Bnl,
the mappings
where n = 1, 2, ... , denote a sequence of domains contained in the z-plane; Suppose that a given point z 0 belongs to each of the domains Bn • We define the
ship
kernel of this sequence with respect to z 0 as the largest domain B containing the point z
,=
IB 1 converge s 71
,,=
Fn(z) converges to the kernel B'. Consequently, the sequence
to a nondegenerate kernel B connected with B' by the relation
['(0)/'. The conclusion on the nature of the mapping' = [(z) follows.
Proof of the sufficiency. Suppose that Bn
z 0 and having the property that an arbitrary closed domain contained in it belongs to every Bn from some n on. Convergence to a kernel is defined =
Bn under a fractional-linear mapping of the z-plane is the image of the kernel B (of course, wid~ respect to the image of the point z 0)' This does not violate con
F (z) = n
2, ... and then let Pn approach 1, the
(+)
converge s uniformly in the interior of the domain A' to a univalent function F (z)
kernel of the sequence IBnl remains unchanged and convergence to it is not
such that F (00)
=
00 and F '(00)
=
1. But then the sequence
'/,-,,;
violated.
I
We shall now prove a convergence theorem. Theorem 2. Suppose that a point z
I~ (0) fn
vergence to the kernel. Furthermore, if we perform similarity transformations
= 1,
=
B. Then, in the same notation,
the sequence {B;I has the kernel B' and it converges to that kernel. Consequently the sequence {Fn(z)l defined by
just as before. Obviously, the kernel of the sequence of images of the domains
z ' = p. z on the domains B71 for n 71
- ....
,.I .
z 0 belongs to each o[ a sequence o[
t,t'i,!
f n (z) = I~ (0) F n (;)
Un (z)1
defined by
234
V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS
§6. MAPPING ONTO CIRCULAR DOMAINS
converges uniformly in the interior of the domain A to the univalent function
c
f(z)= which maps
Denoting the mapping function by ('" [(z), we then have [(00) '" 00. The func
Kv (v'"
tion [(z) can be extended across the circle
c>o,
F(~)'
235
1, 2, ... , n) into the domain
B v obtained from B by inversion about K v' It maps B 1/ univalently onto the B ~ obtained from B' by inversion about K~. Consequently, the function [(z) maps the circular domain consisting of the domains B, 8 1 , ••• , Bn uni
domain
A onto the kernel B of the sequence of the domains Bn • This com
pletes the proof of the The?tem.
valently onto the circular domain consisting of the domains B', B{, ... , B~. We reason analogously with these exrended domains and continue the process indefi
§6. Univalent mapping of multiply connected domains onto circular domains.
nitely. As a result of such extensions in the z'plane, we obtain a limit domain G,
The continuity method
which the function ('" [(z) maps univalently onto a limit G' in the (-plane. Let
In §§ 2 and 3, when we were solving certain extremal problems, we proved
Z '
theorems on the existence of a univalent mapping of multiply connected domains onto canonical domains. In the case of finitely connected domains, an incompar ably stronger method of proof is the continuity method. The proofs that use this a univalent mapping of finitel y connected domains onto the canonic al domains in question and on a single topological theorem of Brouwer. We shall illustrate this
ii' "
method by proving an important theorem on univalent mapping onto circular
I
of a finite number of inversions about the circles K l' K 2 ,
,Kn- The functions z'", Sk(Z), for k = 1,2,"" exclusive of the function z'", z map the domain B
onto disjoint circular domains B (k), for k = 1, 2, ... , that have no common boundary points and lie each inside one of the circles
points in common (some of the circles may degenerate to points). We first prove
Consider a disk jz -
a uniqueness theorem for circular domains.
aJ < r
covering theorem, the domain
Theorem 1. There exists no more than one [unction that maps a given [initely
connected domain B in the z-plane including
5 (z) denote a fractional linear function composed of an even number of inver
S(z) maps the domain G into itself. Consequently, [(z) is invariant under the group of fractional-linear transformations z'= Sk(Z), for k = 1,2, " ' , composed
method are extremely similar to each other. They are based on the uniqueness of
domains, that is, domains bounded by a finite number of complete circles with no
=
sions about the circles K l' K 2' ..• , K n taken in any order. Then, the function [(S(z)) also maps the domain G onto the domain G' univalently since the function
00
7T
domain in the (.plane, also including 00, and that has the expansion (= z + a 1z- 1 + '" in a neighborhood o[ the point z = 00.
L
K 1, K2 ,
••• ,
Kn •
contained in B. Then, in accordance with Koebe's
B(k)
(for k
=
1, 2, ..• ) contains a disk of radius
(r/ 4) IS'; (a)l. Consequendy, the area of the domain
univalently onto a circular
•••
B(k)
is not less that
(rIS'; (a)I/4)2. Since the sum of the areas of all the domains
B(k
l,
for k
=
1, 2, ... , is finite, we see that the series OJ
~ ISk(a)l~
One can easily see that this theorem is equivalent to
(1 )
k=l
'Theorem 1'. Every [unction that maps a circular domain B ([initely con
nected) univalently onto another circular domain B' is a [ractional-linear [unc
converges. If we now denote by Kk, for k = n + 1, n + 2, ... , all the circles
tion.
that serve as boundaries for the domains Since the two theorems are equivalent, we need only prove Theorem 1 '. Proof. Suppose that the domain B is contained in the z-plane and is
bounded by the circles K 1 , K 2 ,
Kn , that the domain B' is contained in the (·plane and is bounded by circles K {, K ~ , ... , K;, and that the mapping in question maps K v onto K~ (v = 1, 2"", n). Without loss of generality, we may assume that the domains Band B' contain the point 00 and that this point is .,. ,
,J
B(k),
their radii by Rk for k = n + 1, n + 2, '"
for k
=
1, 2, ... , and if we denote
, we can show that the series
OJ
~ Rk
~
i
(2)
k=n+l
converges. To see this, let us delete from the domain G the images of the point
z
=
00
under the mappings z
I",
Sk(Z), k = 1, 2, .... Then all the functions Sk(Z)
are regular and univalent in the remaining portion G * of the domain G. There
mapped into itself.
fore, from the general distortion theorem (see
til !
,1
§ 4 of
Chapter II), we have on
Kv
V. UNIVALENT MAPPING OF MULTIPLY CONNECTED DOMAINS
236 (v
=
1, 2 ... n) for all k = 1, 2,' . . . ,
,
' " •
237
§6. MAPPING ONTO CIRCULAR DOMAINS
' i ; > .
m
0;,.
00
IS;' (z) I ~ M.I Sf. (a) /'
.L; A~(.) < e,
R:(.) ~ e,
.L; .=1
.=1
where Jl v depends on v but not on k or z. Therefore, we get the following expression for the circumference 21T Rk v of the circle Kk v onto which the func tion z' = Sk(Z) maps K v :
21CRk •
•
vergence of the series (2) and (3). Now, for a sufficiently large circle K:
21t1
from which it follows that
J_ \' 1 (z')
21t1
(4)
z
K k(.)
Iz I < R.
But
1 (z') - 1 (z) dz'
\' ~
21ti
~ z' -
z' - z
•
K (·) k
Now let d denote the distance from z to the boundary of the domain G.. It follows from what was just shown that
00
m
k=1
.=1
dz' = _I
z
K k (.)
~ R~~~ M:m·.L; I Si. (a) I 2.
k=n+1
~ z' -
\' I(z') dz'
21t1
• =1
H we square both sides of this inequality and then sum over all k and all v, recalling that each circle K k is the image of a boundary circle of the domain B under exactly one of the functions z' = Sk(Z), we obtain
I
This proves the convergence of the series (2). Analogous conclusions hold for
~ _I \' 1.J 21t1 ~
.=1
m
I(z') dz' ~ ~
z' -
K (.) k
I
Z
---=0
tJ.k(·l
R (.)
d
k
'-
.=1
C'. Consequently, if we denote by Rk the radii of the circles into which the = n
1.J
z' - z
for any point z belonging both to G. and the disk
R" • ~ MR. I Sf. (a) [.
circle s Kk, for k
~ _I
/(z)=_1 } I(z') dz' _
•
n
R,
m
K
co
Izi =
we have, in accordance with Cauchy's formula,
IS;' (z) II dz I ~ M.I Sf. (a) I ~ Idz I ~ M.! Sf. (a) ,. 21CR••
)
where (. is a given positive number. This is always possible by virtue of the con
+ 1, n + 2, ... , are mapped by the function ~ =
~~
r(z) defined
.. !
r !
m
.=1
on G, we conclude that the series
A~(.)
!
m
•
m(·)
O. This mapping is unique.
Theorem 7. Every n-connected domain B contained in the annulus q < that has as two of its boundary continua the circles mapped univalently onto an annulus q' < 1(I
Izi =
1(1 =
1 and
1(I =
Izi
=
Iz 1< 1
q can be
< 1 with cuts along arcs of loga
rithmic spirals of given inclination () in such a way that mapped into
1 and
Iz I =
1 and
Iz I =
q are
q'. This mapping function is unique up to a con
stant factor. We note that Grotzsch succeeded in proving analogous uniqueness theorems
a has the form 1
z-a
+
lX l(Z
-a)+ ...
or
Z+~+ z
for extremely broad classes of canonical domains and at the same time established
... ,
by the continuity method theorems on the existence of a univalent mapping of arbitrary finitely connected domains onto such domains (d. Grotzsch [1935]). We conclude our investigation of questions concerning univalent mappings of
according as a is fin ite or in finite.
Theorem 5. Every n-connected domain B in the z-plane with boundary con tinua Kl' ..• , K n can be mapped univalently onto the (-plane with n cuts along arcs of logarithmic spirals of inclinations ()1' ••• ,
respectively to the radial
()n
1, ... ,n, are mapped respectively into arcs of incl ination OJ), given points a and b in B are mapped into 0 and"" respectively, and the expansion of the mapping function about z = b has the form directions in such a way that the continua K v , for
z 1 b+lXo+1X1(Z-b)+ ...
or
J.I =
z+cx.o+ :' +
multiply connected domains with a consideration of the possibility of univalent mapping of any two given n-connected domains onto each other. In the case of simply connected domains, Riemann's theorem tells us that, with very few excep tions, this is always possible. The situation is different in the case of n-con nected domains, where n
> 1.
For it to be possible to map two such domains uni
vaLently onto each other, it is obviously necessary and sufficient that there exist two circular domains onto which they can each be mapped univalently and which can be univalently mapped one onto the other. We can take these circular domains
... ,
of a "special" form, specifically, annuli of the form q
2 m u s t " $ :
n
IS
be the same for the two domains. In the second case, an analogous situation holds
x i =
for one of the circular domains in question and for the domain obtained from the other by the transformation z I =
II z.
Conversely, if these requirements are satis
1JAJXY\
i = I, 2, .•• , n, Aj
j=O
is called an n-dimensional simplex and the points x(k), k
I.
for it to be possible to map two n-connected circular domains of a special form
are single-valued continuous functions of xI' ••. , x n •
2, one, and
0, 1, ...
=
,n,
are
The numbers Ai in (1)
To see this, remember that, since the points x(k), k = 0, 1, ... ,n, do not
> 2, 3n - 6 suitable real quantities determining one of these domains
in the case n
(1)
j=O
called its vertic es. 1) Let us denote this simplex by
=
1JA j = I,
?:' 0,
fied, the two domains can obviously be mapped univalently onto each other. Thus, onto each other, it is necessary and sufficient that, in the case n
x n ) whose coordinates
are determined from the formulas
lie in a single plane, the determinant
coincide with real quantities determining the other. These quantities are called the moduli of the circular domains in question and of all the domains that are
(0)
mapped univalently onto them.
\0)
x l ' ... , X n '
" '/-
(l) (1) Xl , ... , X n '
~
i-'·~
H we partition all n-connected domains into classes, putting into a single (n)
XI ' ....
class all domains that can be mapped univalendy onto each other, then each such class will contain circular domains of a special form and, ip accordance with what
=
2 on one and in the case n
system of n + 1 linear equations in the Ai that has a unique solution. Con
> 2 on 3n - 6 real parameters.
sequently, the Ai are obtained as linear functions of the coordinates i
§7. Proof of Brouwer's theorem 1) To prove Brouwer's theorem as stated in the preceding section,
= 1, ...
,n; hence, they are continuous functions of the
Xi'
Xi,
for
The set of points
of the simplex I that are obtained from (1) when one of the Ai is equal to 0 is w~
need some
called an (n - I)-dimensional face or simply a face of the simplex
additional information regarding the space Rn • A set of points x E Rn with coordinates x
I
cannot be equal to O. This means that the system of n + 1 equations in (1) is a
was said above, all classes of n-connected domains constitute a set depending in the case n
(n) Xn ,
I.
In general,
the set of points of the simplex I that are obtained from (1) if k of the numbers l ' . . . ,x n
satisfying a given linear
Ai are equal to 0 is called an (n - k )-dimensional face.
equation
An (n - k )-dimensional face of a simplex I is itself a simplex in (n - k)
+ asxs +...+ anxn = 0,
a1x1
dimensional Euclidean space, specifically the (n - k)-dimensional plane passing through those vertices of the simplex I that lie on that face. The point x E I
where
Ql'
a 2,
•• ,
,an are not all zero, is called an (n - I)-dimensional plane or
simply the plane Rn -
1•
obtained from (1) for AD = Al = ... = An = I/(n + 1) is called the center of the
The set of all points common to two (n - I)-dimensional
simplex
planes the left-hand members of the equations of which are linearly independent constitutes an (n - 2)-dimensional plane in R n • In general, the set of all points common to k planes under the same condition constitutes an (n - k)-dimensional
,.
""i"
¥
"~"'~~;r
l
I.
By virtue of the continuity of the Ai as functions of xl' " ' , x n
(which we showed above), it follows that an arbitrary point (1) for all nonzero Ai is an interior point of the simplex
X
E
I. obtained
from
I.
Let us show that a simplex I can be partitioned into a finite number of . ',"i"'I;J" .,'~l; I:· plane in Rn • An (n - k )-dimensional plane in Rn can be regarded as an (n - k) simplexes the diameters of which (the diameter of a figure being the greatest :'~~' dimensional Euclidean space. distance between any two points of it) is less than an arbitrary number f >0 in '~: ~ ;',',7); Let x(k)(x(k), ••• ,x(k)), k = 0, 1,"" n, denote n + 1 points in Rn that 1
1) (Sperner [I928].)
n
I~.
, 7/0
I
1) In the case n = 1, the simplex is a line segment; for n = 2, it is a triangle; for n = 3, it is a tetrahedron.
246
§7. PROOF OF BROUWER'S THEOREM
V. UNIVALENT MAPPING OF MUL TIPLY CONNECTED DOMAINS
eO).
such a way that, if two subsimplexes have points in common, then the set of all
point
common points is a common face of some dimension or else a vertex of these
~J'
simplexes. Such a partition can be made as follows, for example: let us look at
of the numbers
all possible permutations
io' il"'" in
... J'
'n
Let us now look at the question of the dimensions of the simplex ,
keeping the notation given above for it. Let d denote the greatest
!xV) - xV ')1
of the numbers 0, 1"", n. Let us
denote by ~iO,il'''',in the set of points x € with Aio SAil:S ...
J'
0' l'
I '"I
.(0)
~ that have the representation (1)
:s \n' .Let us show that
(j'l
-Xl
for i
I--n+l"':' I "\'1 «}) (j'l XI -Xl )
~iO,il'''',in is a simplex. Let
(since one of the numbers xli) for , - x~i'1, , Therefore, if
X
(xl' " ' , x n } €
n
xi·) n -k + l'
elk) .=k I =-~--
i-, l " ' , n.
le}Ol-xII=
(2)
x/v xi
~ and
I! n
Xi
=(n--~+l)el')-(n-'1)el'+l), -"I
nd " ~n+l' J =0, .. " n,
i=
0,1, " ' , n, must be equal to 0).
= ~7=oAiX~i),
i
= 1",', n,
then
n
I d Ai(e~O)-xY'l) ~n+l ~
nd
A;=n+l'
/=0
In particular, for points ~ (k 1, k = 1, 2, ...
(jnl_dn)
0, " ' , n.We then have
I
i=O
From (2), we have (. )
n
I
i, i'=
1,"', nand
=
1=0
~ ( k) (~ik), .•. , ~ ~k)} (k = 0, 1,'" , n) d~note the center of a!1 (n - k}-dimensional ~ . h' (J k ) . I ,x (Ik + I) , ... , x(J n ). Th en, f ace 0 f t h e simp ex ..:. wIt vertIces at x
~
247
,n,
we have
I e!O)-e(k)I~~ , '~n+l' '1=0, I, ... , n-l, Reasoning analogously with the (n - I}-dimensional face of the simplex
•
~i o,i 1"" ,i n with vertices at ~ (l),
Consequently,
••• ,
~ (n), we can show that
_n_d 0,
d.
then the simplex ak has coordinates (0, 1, .•. , n - 1, a), where 0::; a ::; n. Here,
(3)
.;;' .if a = n, we obviously have tk = 1. On the other hand, if a
< n,
then tk = 2
It follows that the diameters of all the subsimplexes after a suffic ient number p
~Cibecause the coordinates of all n + 1 faces of the simplex ak are obtained from
of stages in this partitioning process will not exceed an arbitrary given number
,
f>
0. This is the desired partition of the simplex
1.
(0, 1, .•. ; n - 1) only by deleting the number a when encountered twice. Let e denote the number of all simplexe s
To prove Brouwer's theorem, we need first to prove two lemmas. Lemma L
Let
1
(0, 1, .•. , n - 1, a) by discarding a single number and we obtain the numbers
denote the number of all ak for which tk
denote an n-dimensional simplex with vertices xU) [or
Ej , [or.i
(3)
2. Then, from what was said above,
~ tk=e+2j.
(4)
k.=1
0, 1, "', n, in such a way that the [ollowing three conditions are satis[ied: (1) each point x € 1 belongs to at least one o[ the Ej ; (2) x(j) €
Ej ;
=
p
j = 0, 1, ... ,n. Suppose that the points o[ 1 are distributed among n + 1 closed sets
ak with coordinate s (0, 1, .•. , n) and let [
=
On the other hand, if a face of a subsimplex ak lying inside
belongs to two of the simplexes ak. Let g denote the number of such faces
[or k Ie- j. Then, the sets Ej' [or j = 0, 1, ... ,n, have at Ie ast one common point.
1
belonging to distinct ak' A face of a simplex ak lying on one of the faces of the
into a finite number of subsimplexes
simplex
aI' a2' ... , ap in such a way that any two of these simplexes have in common
< v2 < ... < v q ,
intersections
simplexes. Consequently,
sn
nE
j,
j
=
0, 1, " ' ,
n -
1, that satisfy
Sn
will have an odd number h of faces
Sk
with
coordinates (0, 1, ... , n - 1). Thus, we have
p
assign the n numbers that are the coordinates of the vertices of ak in it. Let us
~ tk=2g+h.
k=1
show that the number of simplexes ak of which all the coordinates are distinct 1 because then
is an In - I)-dimensional simplex the points of
Our assertion is considered proved for the case of (n - I)-dimensional
let us assign to it the
n + 1 numbers which are the coordinates of its vertices and to each face let us
=
n
sn
condi tions (1), (2), and (3) of the Ie mma.
number VI as coordinate. To the simplex ak let us assign as coordinates the
is an odd number. This assertion is true in the case n
But
Sn'
which belong to the
Ej for j = 0, 1, ... ,n. Each
vertex of the simplex ak belongs to one or more of the sets Ej • H it belongs to the sets E VI' E v2' ••• , E vq ' where VI
1, on the other hand, will have coordinates (0, 1, ••. , n - 1) only if
this face lies on
only faces of some dimension or other. Let us show that among the ak there exists a simplex containing points of each set
has coordinates
(0, 1, ... , n - 1), it appears twice in the sum l~=Itk because each such face
Ej does not contain a single point on the [ace s j with vertices in x where f.:s: n. On the other hand, x belongs to certain of the sets Fk, let us say to F l' F 2 " ' " Fg , where g ~ h. The sets FI"'" Fg are contained in the ball of radius (with center at x. But, from our construction, this ball con :1,
tains points belonging to no more than n of the original sets E k, let us say to the sets E l' E 2' ••. , Eh , where f.:s: h ..:s: n. From 2), the sets F l' .•• , Fg are combined with certain of the sets E h
••• ,
Eh' Consequently, a point xES can
belong to no more than n of the modified sets Ek , namely to certain of these sets that contain the original sets E I' ... , Eh' Thus, after we adjoin the sets Fk to the sets E k , every point xES belongs to no more than n of the sets E k • Since a is a boundary point of the set E, the simplex 1 contains an interior
\"'
point c that does not belong to E. Let us distribute the points of the sitnplex 1 and heJlce let us redistr ibute the points of the set K
=
E U1 among the sets Ek E v2' .•• , E v p ' where
as follows: If a point xES belongs to the sets E vI' p .:s: n, then the entire segment cx (one-dimensional simplex) belongs to all these sets and only to them. With this new covering of the set E U 1, the point c is the only point that can belong to more than n of the sets Ek • But c does not belong to E. Consequently, E does not contain points belonging to more than n of the sets E k • This completes the proof of the lemma. Brouwer's theorem. A topological mapping of a set E C R ll onto a set E'
c
R~ maps interior points of E into interior points of E •
Proof. Let a denote an interior point of the set E and let 1 denote ail arbi trarily small simplex contained in E and including a as an interior point. Corre
E', and corresponding to the simplex I 1'. Let us subdivide 1 into n + 1 closed subsets Ej • j = 0, 1, ... ,n, in such a way that
sponding to the point a is a point a' in is a set
each Ej
l' c E'.
Let us show that a' is an interior point of the set
contains only one vertex and no point of the opposite face and in such
a way that a is the only point belonging to all the sets Ej • The mapping of E
,.,
I
···"---r .-
--------.-T-.i'q.. -'·-~~ - T---
~~~#iI~.--.--_._--
f-:-'-~C
. -Ii .-
~
.... "".
--~.;:;;:;L:;iii;
255
§1. MAPPING OF A MULTlH..YCONNECTED DOMAIN
without loss of generality that case, the function t
=
c:
and b are finite and let us observe that, in this
log «z - a)/(z - b)) can be extended into B along an arbi
trary path and it maps the domain B onto the t-plane with the point The inverse function z
CHAPTER VI
=
excluded.
00
¢(t) is regular in that plane. Therefore, if there existed
a function (= f(z) that maps the domain B conformally onto the disk function (= f(¢ (t)) would be regular in
1t I
m > a on the circle 1(- al = O. Let (3 denote an arbitrary number in the domain 0< I( - aj < m. Since 5(() - (3 = (5(() - a) + (a- (3), it follows from Rouche's theorem that the function 5 (() - {3 has a zero y in the disk I' - al < O. Here, y is distinct from a and (3. It follows from 5 (y)
=
=
can be nODe in the domain
1'1 > 1.
We proceed analogously in the case in which 5 is a parabolic transformation
(' =
=
that the transformation ("
DO
1(\ =
T5T- I (() is a transformation with unique fixed point 00. Therefore, (+ a and, consequently, T5(() = T(() + a, where a = const. This shows
that ('
1(\ < 1 and hence that there
Therefore, all the fixed points lie on the circle
§2.
with fixed point (I' By making the transformation t = T (() = 1/(( - (1) we see
¢(() maps a sufficiently small neighborhood of the
fixed points of the transfonnations 5 € G in the open disk
converge as
to the constants (I and (2' This convergence is uniform on an arbi
00
be necessary for
{3 that ¢ (y ) = ¢ ((3).
point (= a univalently. This contradiction leads to the conclusion that there can be
+
(
trary closed set not including the points z 1 and z 2. This last conclusion will
But, for sufficiently small 0 and m, this last cannot be since ¢' (a) ~ O. Conse quently, the function z
-->
n
=
5( is obtained from the equation 1 1 ~'_~I = t-'I
+a,
1.
.~
i'
(3)
i!{ajW;:;;1ffi)j¢j"'MM:t"~Wi\t":~'4it;*'~t-,j;¥"\l'~'t,;;;,,-\t,,,.,~,,-\:;,,,\,:",, "i:Vi~,¥-'U!i:;i'i.-;i~~;I,£{~':;Eiltj;'~"'\#"':"~.J;}.l:jfu~~m::c'* ":rlt:';:2f'if-'y,r~¥Jrti1i~'gt"(~~~"'"i.l)V',"1z'ji""J'l.J.:!j!@i"
260
VI. MAPPING ONTO A DISK
§I.
where a 1= 0 because, in the opposite case, we would have (' impossible. On the other hand, the transformation (/ is defined by the equation
1
C' - C1
=
1
C- C1
=
which is
= (,
sn( (for n
=
261
point equivalent to it and cannot be reduced to a point. Thus, for us to proceed
±l, ±2,"')
in B along a closed path with some element of the function fez) and return to the original point with the same element of fez), it is necessary and sufficient that it be possible to reduce this path by a continuous deformation to a point without
+ na.
(4)
getting outside B. I) In conclusion, let us pause to construct a Riemann surface connected with
Therefore, once again, it eannot be the identity transformation for any n. We note
the mapping that we have been considering. We confine ourselves to the case of
that the sequences of the functions sn( and s-n( converge uniformly to (Ion an arbitrary closed set including the point (I as n
-->
a finitely connected domain B.
+00.
If the domain B is n-connected, we can render it simply connected by means
This property shows, in particular, that G is an infinite group, that is, that
of n - 1 suitable transverse cuts (broken lines). We then take an infinite set of
it contains an infinite number of distinct transformations and that fez) is an infi
copies of the domain B so cut and consider them as lying one above another. We
nitely-many-valued function in B. On the other hand, the function ¢«() assumes
choose anyone of them as the basic domain and denote it by B o. We adjoin to
1(' < 1 every value in B at an infinite number of points. All points in the disk 1'1 < 1 at which the function ¢ «() assumes the same value are called equivalent
in
Bo along each edge of each cut one of the remaining copies. We make this con nection-this gluing-along opposite edges of the cuts. As a result, we obtain
points. For every point a in the disk 1 a 1 < 1 there is an infinite set of points equivalent to a. These are all obtained from each other by transformations of the group G. Since these points are zeros of the function ¢ «() - ¢ (a), they do not have points of accumulation in 1(I < 1. It follows, in particular, that the group G
a (2n - I)-sheeted surface B 1 lying over the domain' B and having (2n - 3) (m - 2) free edges of cuts, that is, edges along which no gluing has been made. Then, we again adjoin along each of these edges one of the as yet unused copies of the cut domain B in such a way that we again perform the gluing along opposite edges of
is countable.
the cuts. We then obtain a new surface B 2 with a finite Dumber of sheets and
Since the function fez) is not single-valued in B, the question arises as to
some number of free edges of cuts. If we continue this way indefinitely, we finally
what conditions must be satisfied by the path l contained in B and issuing from
obtain an infinitely-sheeted Riemann surface Boo.
and returning to a point z 0 in such a way that as we proceed along it with some element of the function
MAPPING OF A MULTIPLY CONNECTED DOMAIN
f (z)
If we consider the initial element fo (z) of the mapping function fez) defined
we return to the point z 0 with the same element.
Consider the case in which B does not include
00.
in the domain B 0, the analytic continuation of this element into B is identical
If, as we proceed along l from
to its analytic continuation onto the surface Boo. As the result of all possible
z 0 with some element of fez), we return to z 0 with the same element, the corre sponding point (= fez) issuing from some point (0 returns to (0 after describing a closed path l' in the disk 1(I < 1. The path l' can, by a continuous (for example. similarity) deformation be reduced to a· point that does not leave the disk 1 (I < 1.
analytic continuations of fez) onto the surface B oe , we obviously obtain a single valued function on that surface. Consequently, the function (= fez) that we studied above also provides a one-to-one mapping of the surface Boe onto the disk
Obviously, under this deformation, the preimage l as it is continuously deformed
1(I < 1.
The surface Boe is called the universal covering surface of the
domain B. The modular surface constructed earlier (see §6 of Chapter II) is, in
also reduces to a point without getting outside B. Let us show that, conversely,
this terminology, the universal covering surface of the domain consisting of the
if the path l can, by a continuous deformation, be reduced to a point without ever
plane with the points 0, 1 and "" excluded.
getting outside B, then, by proceeding along it with an arbitrary element of fez), we always return to the original point with the same element of fez). This is true because, otherwise, under such a deformation, the corresponding path l', While being continuously deformed in ,(,
< 1,
will go from the point (0 to some other I) If B includes
j
00,
an analogous condition must be satisfied on the Riemann sphere.
262
~.
-
MAPpmG =TO
=: . --",'----::=::0: OOONDAro~ OF
§2. Correspondence of boundaries under a mapping of a multiplyO~
connected domain onto a disk~,;
In
§ 1,
,,~~
we examined a function fez) that maps a given multiply connected
domain B conformally onto the disk
I"
< 1 only inside B. If we now let the
to
point z approach the boundary of the domain B, we can easily show that the sequence of the correspon~ling points' approaches the circle In other words, for arbitrary from the point z
c
I:
1'1
";'1 uniformly.
.,
~
> 0, there exists a 0> 0 such that, if the distance
corresponding point 'from the circle
cB
I"
=
1 is less than
f.
'2,
'0,
0, the curve A is constricted to the point
'0
Conse
Then we need merely take for 0 the distance from the set E to the boundary of
quently, as we let z approach a, the branch fo (z) approaches
the domain B. When we make a more thorough study of the behavior of the function
we approach a with another branch 5fo (z) of the function fez), the corresponding
fez) as z approaches the boundary of the domain B, we shall confine ourselves
point' = 5fo (z) approaches the point 5'0' Thus, as we let z approach a, the
to the case of a finitely connected domain B. Such a domain can easily
be mapped
uniformly. 1£
only limiting values of the function fez) can be the points 5,0' where 1'0
1=
1
and 5 c G. When the entire boundary of the domain B consists of isolated points, it fol
univalently onto a domain bounded by closed analytic Jordan curves and isolated points. The correspondence of the boundaries under such a mapping has already been studied. Therefore, for our purposes, it will be sufficient in the future to
lows from what was said above that, as we let z approach the boundary of the
consider a domain B with a boundary consisting of closed analytic Jordan curves and isolated points.
domain B, the function fez) can have as its limiting values only values belong ing to a countable set of numbers, which are points on = 1. It follows that, as
Let a denote an isolated boundary point of the domain B, which we may assume to be finite. Suppose that the closed disk
Iz _ a I ~ r
, describes a path in the disk
Iz _ a 1= r
one time in the positive direction, the branch fo(z) of the function fez) becomes
that terminates on the circle
some point different from these points, the corresponding point z
contains no boundary
points of the domain B other than a. As we go around the circle I:
"I < 1
I"
=
1'1 =
1 at
;•.-i;,,;,;;,;,:;;:,;,.:·;,,;;o;;'''''"~~·'''I'·.~.
266
VI. MAPPING ONTO A DISK
"""'-:;;;.....;="-,,,...
=_-""'-,""'"
267
with either sign chosen is then regular and bounded in the neighborhood in ques
these arcs). Consequently, the function s «() is invariant under transformations
I"
..· ---":;;";,,,;.,,;,;;'';'~':;.,;; --
§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION
of the group G. This last shows that, if we use the function z disk
"="""".~c,-"""""",,,",,,,,,,,,,,,",,;:';''';;,~;:;:',;;,-.;:::,.~
=
tion of the point z = a, and, with suitable choice of the value for F (a), regular
¢ «() to map the
at the point a itself.
< 1 onto the domain B, we obtain in B a single-valued bounded harmonic
Obviously, F(a) is nonzero. Therefore, the function u + iv
function is identically equal to unity in B because, otherwise, it either would
that a function u(z) that is harmonic and bounded in a domain B is harmonic at
min~mum inside B or would approach an extreme value as z approaches isolated boundary points, which, in accordance with the maximum
points.
principle and the beginning of the present section, is impossible.
I(I < 1.
00
S
~
«(t) (~k»)
de =
\~
21t
k=1
so that Lk=llength
The maximum principle of harmonic functions has an important generalization: tIf a function u(z) is harmonic and bounded above in a domain B (that is, its
But
values in B do not exceed some finite constant) and if all its limiting values, as
OJ
(0) = '\'1 ~
~ '\'1 length 21t ~
1
its argument approaches any except finitely many boundary points of B through
fr(k)C{k) \'0 1 2 ,
values in B, are no greater than M, u (z) ~ M everywhere in B. An analogous
"=1
C(k)c(k) 2
=
all isolated boundary points when it is extended in a suitable manner to these
attain a maximum or a
Consequently, s «() '" 1 in the disk
=
a and the function u(z) is harmonic at z
a. It follows
function s (f(z)) that is equal to unity at all nonisolated boundary points. Such a
is also regular at z
generalization holds for functions bounded below.
277. This completes the proof.
=
Proof. We may assume that the domain B includes the point
00.
Let al" ..
.. , an denote the exceptional points referred to and let H denote the diameter
§3. Dirichlet's problem and Green's function
of the boundary of the domain B. For arbitrary
A finite real function that, together with its first two derivatives, I) is single
f> 0, consider the function
n
valued and continuous in a domain B of the z-plane and satisfies Laplace's
v(z)=M - u (z)
equation in B is said to be harmonic in B. Such a function can be regarded as
+e !tog
I _
H ~,.Il
k=l
the real part of Some analytic f unction that can be extended in B along an arbi
which is harmonic in B. As z approaches boundary points other than al"", an,
trary path. A function is said to be harmonic at a point if it is harmonic in some
this function approaches nonnegative limits since each term in the sum above has
domain including that point. We note that, if u (z) is a harmonic function in a suf
a nonnegative value. As z approaches one of the points ak, this sum approaches
ficiently small neighborhood of a point z
+ 00, but u(z) is bounded above. Consequently, the limit of v(z) as z approaches
=
a with the point a itself deleted and
if it is bounded below in that neighborhood, that is, if u(z) that is, if u (z)
S. M),
it will also be harmonic at z
is extended to include the point z
=
= a
~ m
(or bounded above,
. aoy of these points is + 00. Since v (z) is obviously not a constant, it cannot
if its domain of definition
lJave minima inside the domain B. Therefore, v(z) n
a in a suitable maoner. To see this, let v(z)
ll(z)~M +e !tog~
denote the harmonic conjugate of U(z) and let w denote the increment in the func tion v (z)
as z moves once around a sufficiently small circle
sufficiently large circle
Iz I =
R in case a .
=
Iz - al
= f
~
0 in B; that is,
,.
k=1
(or a
This holds for arbitrary
(0). The function
f
> O. By letting
f
approach 0, we obtain u(z):S M,
which completes the proof.
+ 2" (u+lvj
Let us new pass on to Dirichlet's problem. Suppose that the domain B has a
F(z)=e- '"
boundary
K consisting of a finite number of closed Jordan curves without com
mon points. The Dirichlet problem for such a domain consists in 1) The derivatives here refer only ro derivatives at finite points. In what follows, the harmonic functions mentioned may at rimes be multiple-valued, bur in such cases an arbi trary branch of such a multiple-valued function will be single-valued in an arbitrary suf ficiently small neighborhood of each point in the domain B.
fin~ing
a func
tion u(z) that is harmonic and bounded in B, that is continuous in B except possibly at countably many points on the boundary K; and that assumes prestated
~:j.
·
1 ~
.! .
i
values at points of continuity on the boundary K. The boundary values are subject
268
269
§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION
VI. MAPPING ONTO A DISK
I(I < 1.
If we now return to the z-plane, the function V
«;) becomes
only to the condition that they constitute a function on K that has a finite or
identical in
countable set of points of discontinuity. Let us show that this problem always
the function u(z), which is single-valued and bounded in B and which satisfies
has a unique solution.
the Dirichlet problem posed above. It also follows from what was said above that
I(I < 1,
Let us map the domain B onto the disk
just as in
§ 1.
this problem has a unique solution. We note that if certain of the curves constitut
Consider the
ing the boundary K degenerated into points, the Dirichlet problem for the domain
1(' 1
1 belonging to the class (1) of § 2. Let V «(') denote a function whose values coincide with the values u tz ') at points
set of points (' on the circle
'=
z' of the boundary K that correspond under this mapping to the points ( on I(' I '= 1. This real function V «( ') is defined, bounded, and continuous almost everywhere 1) on the circle
I(
circle the condition V (S «('»)
1 '= '=
1 and satisfies almost everywhere on that
V «(), where S is an arbitrary transformation be
longing to the group G of §2. Thus, we arrive at the problem of finding a function V «() that is harmonic and bounded in the disk \ 'I
< 1 and that coincides with V «() at all points of
continuity of the function V«() on
1'1
'=
1.
obtained would not always have a solution. To see this, note that the function u(z) which provides a solution to the Dirichlet problem for such a domain is, by
I'.. .:. virtue of what was said above, a harmonic function at all isolated boundary points. ."
'~·.Therefore, it is uniquely determined by its values on the remaining boundary ,; curves. This shows that the values of the function u (z) at isolated boundary
'iCf points cannot be chosen arbitrarily. :J:'
Thus, we have proved the existence and uniqueness of the solution to the
i) Dirichlet problem for an arbitrary finitely connected domain B bounded by Jordan
~;curves and for arbitrary given values on the boundary K that constitute either a I.
{(continuous function or a function with a finite or countable set of points of disl;,
Consider the function
!'continuity. To find the solution of an arbitrary Dirichlet problem for a given
2",
1 \
19
U(C)=2~ ~ U(e ),
"
l - rs
.f: domain B, it is sufficient to know the solution of anyone particular Dirichlet
.
fl'.\
I
_.dB, C=re''l',
{problem. Here, it is a question of finding the Green's function. The Green's func where the integral is in the sense of Riemann. By virtue of the kn5>wn properties of a Poisson integral, this function is harmonic and bounded in the shift to
1 (' I
'=
I(I < land,
'if tion for a ~,
under
1, it coincides with V «() at all points of continuity of the
domain B is defined as a real function g (z, () satisfying the following
(three conditions: 1) for every (€
B, it is a harmonic function of z in the domain
. B except at the point z'= (; 2) g(z,() approaches +00 as z --- ( in such a way
1/1 z -
~ remains bounded if ( is finite and in
function V «(i). On the other hand, any function V «() possessing these proper
that the difference g (z, () - log
ties is unique. To see this, suppose that there exists another such function V 1«()'
such a way that the difference g (z, (0) - log I zl is bounded if .' '= 08; 3) as () approaches the boundary K, the function g(z, () approaches O. In accord
Then, for arbitrary ( in
1'1
< 1, we have, in accordance with Poisson's formula
210
1 \
UI (C) = 2~ ~ UI
('9)
Re
RS - r
B
RB-2Rrcos(6 _ 'f')
+ rB
dB C_ ,
-
ilnce with the maximum principle, g(z, () must be positive everywhere in B. If
l; that is, this difference is a solution of a particular Dirich let problem for the domain B. Conversely, if u(z) is a solution of this Dirichlet problem, then the function log
(1/1 z - 'I), + u (z)
is obviously a Green's function
transformations of the group G. In fact, the functions V(S«()) and V«() are
for the domain B. An analogous situation holds for
both solutions of the problem that we have just examined and hence they are
sheeted mapping of the domain B by means of a function z*
1) In the present case, with the possible exception of a countable set of points on
I ~I = 1.
(
is harmonic in B and continuous in B and assumes on the boundary K the
('=
00. Under a single
[(z) onto another domain B* bounded.by Jordan curves, a Green's function g(z, () is transformed into the function g([-l(z*), [-1«(*»), which, obviously, is a Green's function for the domain B*. This shows that, if we know a Green's function for one domain, '=
270
VI. MAPPING ONTO A DISK
§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION
we can, by means of univalent mappings, obtain Green's functions for various
It is obviously equal to 0 at points on the real axis and it coincides with u (t) on
other domains of the same order of connectedness. In the case of a simply con nected domain B, there is no Green's function, as the function cates, where t
It 1
(0), it follows that, as we move along the bound
the domain B lying inside y consists of simply connected domains. We denote
"!aryof B in a neighborhood of zoo the function h(z, ,) always changes in the
by B 0 whichever one of these domains has the point Zo on its boundary. Let us
,rsame direction. More precisely, as z describes the boundary of the domain B in
map B 0 onto the halfplane t
=
0 (t) > 0
in such a way that z
O. If we denote the mapping function by z
regular in the domain
o(t)
~
=
on
t
1
=
tthe positive direction, h(z, ,) increases. Now, if Zo is an interior point of an analytic arc on the boundary of B, then the function ¢ (t) is regular at the point
and its real part u (t) is also continuous in
0 and is nonnegative. Noting that u (t)
ficiently close to t u (t)
0 (t) > 0
=
z 0 is mapped into ¢ (t), the function F (¢ (t)) is =
=
= O. But then, the function
0 at points on the real axis suf
0, we describe a circle It \
p sufficiently small that
=
0 on the diameter lying on the real axis. Let us now define a function U(t) I
=
p as follows: U (t) coincides with the function u (t) on the upper semi
circle and it assumes at points in the lower semicircle the same values as the function u (t) assumes at the conjugate points in the upper halfplane with with oppos ite sign. The function U (t) is obviously continuous on the circle
I" =
p•
Consider the integral
,)
is regular at
the point z o'
'g,
By analogous reasoning, we can obtain a ~.,' functions:
g~neralized
theorem on harmonic
Let u(z) denote a function that is harmonic in a domain B bounded by closed ~aTtalytic Jordan curves. Suppose that u (z) is continuous in B and that its values On
the boundary of B constitute an analytic function of a parameter determining
points on the boundary (that is, a function that can be represented in a neighbor hood of each boundary point in the form of a series of powers of the parameter in
2"
1 \ '0 U(t)=2'IC ~ U(pe l ) .
F (z) and hence the function p (z,
pi ,,
rl It,
_.\
I
_.de,
,
t=re'l'.
o It defines a function U (t) that is harmonic in
question). Then an analytic function F(z) the real part of which is u(z) is regular in B.
I t I < P and continuous onl t I :$ p.
Proof. Just as above, we shift to the t-plane. Then, the values of u (z) on
the boundary of the domain B in a neighborhood of a point Zo can be represented
conclude that the last integral approaches 0 as
by a Maclaurin series of powers of the real parameter t. In the complex t-plane,
integral, we have
this series defines a function that is reguiar at t
=
273
§3. DIRICHLET'S PROBLEM AND GREEN'S FUNCTION
VI. MAPPING ONTO A DISK
272
0 . With regard to the middle
f -->
0 and real on the'real axis
I
(in a neighborhood of t = 0). When we return to the z-plane, this fast function be
._0
lim
comes a function 'I'(z) that is regular at the point Zo and equal to u(z) on the boundary of the domain B. in a neighborhood of the point z o' The difference
\ ~
\z-CI=I
d
dg
U •
ds~lim HO,,_t'~ _.
d log
U
I z - CI ds d.
-lim
(
d.
U1,_CI
~-2
O. we obtain
1 \
u(q=2~
dg k u(z) dn ds.
(2)
regular at the point z o· Let us now show how, once we know a Green's function for the domain B, we
This formula is known as "Green's formula". It expresses the values of the func
can find the solution of an arbitrary Dirichlet problem for that domain.
tion u(z) inside the domain B in terms of its values on the boundary of B. Since
Let us first consider the case of a domain
closed analytic curves. Let u(z) denote a solution of a Dirichlet problem for the domain
the function p(z, ()
B with boundary K consisting of
B with given boundary values which define on K an analytic function of
=
g - ih is regular on K and hence dg/dn
=
dh/ds (one of
the Cauchy-Riemann conditions), we can put the preceding formula in a different , form:
u (q =
a real parameter determining points on K. Green's formula I)
2~ ~
rt (z)
dh (z,
q.
(3)
K
\ \ ,
\ ( dU l
~ ~ (Uarzl - Ulau) dx dy = -
~ u dn
B.
Let us show that formula (3) holds for a finitely connected domain B bounded
1(.
is applicable to the functions u(z) and Ul(z)
o~tained
dU)
Ul dn ds
-
by arbitrary closed Jordan curves and for an arbitrary given function u (z) on the
g(z, () and to the domain BE from B by subtracting the disk \ z - (\ < (.2) Here, the integral on the
left is over the domain
4i
boundary K, where the integral on the right should be understood as a Stieltjes
=
integral. Let g(z,. () denote a Green's function for the domain B and let S> 0
that on the right is over the boundary K E of the domain
denote a number less than the distance from any of the zeros of the function
BE in the positive direction; ds denotes an element of arc on KE; and d/dn de
p;(z, () (where p
notes the derivative with respect to the inner normal. Since Au
>'balf the distance between different boundary curves constituting K. On the set
and since u 1
=0
=
Au 1 = 0 in BE
on K, the preceding formula reduces to ' } ,
dg U dn ds
+
\
~ Iz-CI=I
dg
U dn
ds -
~
du g dn ds
=
g - ih) or the point (to the boundary K and also less than
consisting of all points of the domain B whose distance from the boundary K is
= O.
/) or greater, the function g(z, () has a positive minimum
(1)
K A of points of the domain B defined by the equation g(z, ()
Iz- 1=1
= (,
Consider the set =
>.., where 0
.. of the set K A consisting of tion z
=
p; I) This formula is also valid for a domain B including z = De, as we can see if we apply it first to the domain BE obtained from B by subtracting the domain \ > II E, let E approach 0, and then proceed as described in the text. 2) For finite ~. In the case ~ =00, we take the domain \ > II E.
zl
zl
points at a distance no greater than [) from a given boundary curve L of the do main B consists of a single closed analytic Jordan curve surrounding L because,
",,_."_::- _"_"
._~".
274
"'..__.
::,..~
..__
. -, 0, the values of the func 0'
=
= -
lim g «, z)
s 0;
that is,
d(z, () SO for every ( in B. On the other hand, as we let (approach K, we limd (z, ()
=
lim-g (z, ()
?: 0; that is d (z, () ?: 0 in B. Consequently,
; d(z, () = 0, that is g(z, () = g(' z) for every z and ( in B.
f.
Consequently, the integral in the right-hand member of (4) approaches the integral
§4. Application to a univalent mapping of multiply connected domains
over K as ,\ --. O. By taking the limit, we obtain
We can use the solution of the Dirichlet problem to give new proofs of theo
1l(C)=2~) U (zo (0)) do,
, rems on the existence of a univalent mapping of finitely connected domains onto
i; certain
that is,
canonical domains, for example, onto the plane with rectilinear parallel
!
U
(C) = 2'1t ~
i,
U
(5)
(z) dh (z, C).
Finally, formula (5) remains valid when the function u(z) has a finite number of points of discontinuity on the boundary K but remains bounded
(I u (z)1
< M) in
B. To see this, let us include the points of discontinuity on K in the interiors of small arcs (z(O'~), z(O'k)), suchthatIIO'~ -O'k\
cuts or onto the plane with cuts along arcs of logarithmic spirals of equal incli nation. Let us stop here to prove Theorem l' of §2 of Chapter V on the mapping of an n-connected domain B in the z-plane onto the (-plane with finite parallel straight-line cuts of inclination () to the real axis. As we know, it will be suf ficient to consider the case when the domain B does not include z
=
00
and is
bounded by closed analytic Jordan curves K 1> K 2 , •• " Kn , where K n is assumed to be the external boundary.
lll(z).(o))-u(zo(o))lda.
K
But off these arcs, u(z,\(O'))
~ll
u(zo(O')) uniformly. Consequently, for sufficiently
Consider n - I functions uk (z), for k
=
1, ... , n - 1, that are harmonic in B.
Uk (z) is equal to I on K k and equal to zero on all K k , for k' ~ k. For each k, let vdz) denote the harmonic conjugate of uk(Z). The func
Suppose that each
small '\, the last integral is also less than d2. This proves formula (5) in the
tions Vk(z) are in Eeneral multiple-valued in B but, in accordance with §3, they
present case.
are continuous on B. Let us denote by wk,l the increment in the function Vk(Z)
276
277
§5. MAPPING OF AN n-CONNECTED DOMAIN
VI. MAPPING ONTO A DISK
as z moves around K I in the positive direction (with respect to the domain B).
cannot be the case. Thus, the univalence of fez) in B is proved and so is the
One can easily show that the determinant IWk,l1 of the numbers wk,l' is nonzero.
fact that the image of the domain
B under the function (= fez) is a plane with
e with the
real axis. The function f(z) + c
Let us suppose, to the contrary, that this determinant is zero. Then there exist
parallel rectilinear cuts of inclination
real constants ~k , for k = 1, •. " n - 1, not all equal to zero, such that
yields the same mapping, and for some c we have, in a neighborhood of z
I~~lAkwk.1 = 0, l = 1"" , n - 1. The function
the expansion I
n-l
z_ a
/(z) = ~Ak(Uk+I'Vk)' which can be extended into
B along an arbitrary path, returnS to its original
f'~
value as z moves around each of the curves K I encircling K n • But then, this function is single-valued in B and hence is regular in B. The real part of the
_~(e-iBI(z - a» on the boundary K
=
(z - a)
+..., 08.
By analogous reasoning, we can prove Theorem I' of §3 of Chapter V re along arcs of logarithmic spirals of equal inclination. §5. Mapping of an n-connected domain onto an n-sheeted disk
fez) == const in B. But this is impossible since the real part of the function fez) vanishes on the curve Kn and is nonzero on at least one of the curves Kk for
Iw k,ll ;, o.
Now, suppose that uo is a harmonic function in
(XI
a,
8'Uding the mapping of a finitely connected domain B onto a plane with cuts
function fez) is constant on each of the curves K k , for k = 1, ... , n. Then, reasoning as in the proof of Theorem 2 of §2 of Chapter V, we can show that
k = 1, ... , n - 1. This contradiction proves that
+
is finite and the expansion z + a1z- 1 + •• : 1£ a =
k=1
=
In conclusion, let us look at the possibility of mapping an n-connected onesheeted domain
B onto an n-sheeted disk
I(I
< 1. By an n-sheeted disk, we
. ,mean here a Riemann surface lying entirely over the disk
B that is equal to
I(I < 1
and having over
each point of that disk exactly n points (a branch point of order p is considered
U~=IKk and that Vo is its harmonic con
as p points). This mapping is assumed to be effected by a bijective regular func
jugate. Let us choose real constants Ak' k = 1', •.• , n - 1, such that the function
tion. If we do not wish to resort to the concept of a Riemann surface, the mapping
n-l
should be understood in the sense that, if (= f(z) is the function performing the
'Vo+ Ii,",,} ~ Ak'Uk
mapping, then the function fez) - (0 has exactly n zeros in B for arbitrary (0
will be single-valu~d in B. This is possible since IWk,ll;' O. Let us show that
such that 1(0 I < 1 . (with a zero of order m counted m times) and has no zero at
the function
all in n-I
f(z)=z
I
a+ei61(uo+I'Uo+
~Ak(UIi+I'Vk»)
Theorem. Every n-connected domain B in the z-planethat has no isolated
boundary points can be put in one-to-one correspondence with the n-sheeted disk
1i=1
. ,(I < 1 by means of a regular function. If B is bounded by closed Jordan curves
is univalent in B. This function is regular in B except at the point z = a and assumes values on K I (for l = 1," " n) that lie on some straight line d l at an inclination with the real axis. This shows that, if (0 does not lie on the
e
straight lines d l , l = 1, ••. , n, the increase in the function arg (f(z) - (0) as z moves around the boundary of the domain B in the positive direction is equal to Zero. On the other hand, this increase mUSt be equal to 2" times the difference between the number of zeros and the number of poles of the function fez) - (0 in the domain B. Since f(z) - (0 has a single simple pole in B, namely, at z
B for any (0 such that 1(0 \ > 1.
=
a,
~.
tK1'" "
K n, then the function defining this correspondence is continuous in B
t;.and it can be normalized in such a way that, at n given points al"", an, where {, ak € K k , for k = 1, ... ,n, it assumes a single given value a such that \ a\ = 1 ill
'and, at two given points band c of the curve K n that are distinct from the an
and that are distributed along with the an in a definite order on K n , assumes
. given values f3 and y distinc t from a and distributed, along with a, in the same qrder on the circle
1('
= 1. Under this normalization the mapping is unique. (Here,
it follows that fez) assumes the value ~ at exactly one point in B. If the func
"order" refers to the direction of motion around the boundary of the domain
tion fez) assumed any value (1 at more than one point in B, this would also be
the boundary of the n-sheeted disk.)
true of points sufficiently close to (I' which, by virtue of what was shown above,
B and
i;II~I~'~'"'~tl~~Jll!lll1llll'_>!JIl/;Illi·.t~~~~~'';;>~W~~'!ftj!lm l/lN;it1.~..,.
§s.
VI. MAPPING ONTO A DISK
278
Proof. As always, it will be sufficient to consider a domain B that does not include
00
and that is bounded by closed analytic Jordan curves Kk,
k'=
1, ... , n,
where K n is the outer boundary. Suppose that the curve K k (where k,= 1, "', n) is defined by the equation z and that the point
Zk
Instead of the disk
'=
zk(t),
for 0
:s 1.
Suppose also that zk(O)
Kk in the positive direction as
(t) moves along
1'1 < 1_ and the
~ t
= ak
t increases.
n-sheeted disk mentioned in the theorem, we
q
r,_ =7_'
;
IlftiIDWIIll ~~:lI;~
MAPPING OF· AN n-CONNECTED DOMAIN
279
past the real axis and it is a function tbat is regular on that axis except at the point (= O. It follows tbat tbe function f(z) neighborhood of the point z
'=
(l/l7i) log (z - ak) is regular in a
In tbe same way we can prove that the function
= ak.
f(z) + (l/l7i) log (z - b k ) is regular in a neighborbood of tbe point z
=
bk • It fol
lows from what was said above that the function v(z) approacbes +00 uniformly, at the same speed as the function
_77-
1
log
Iz -
as
akl,
z approaches
through
ak
can obviously choose an arbitrary simply connected domain G that can be mapped
values in B, and tbat v(z) approaches
onto tbe disk and the n-sheeted surface over it respectively. For such a domain
approaches bk through values in B. This
floperty enables us to sbow that the function w = f{z) maps the domain B onto
lying over the (w
'=
f{z)
=
function w
'=
u + iv)-plane, we take the strip G: 0 < !lHw) < 1 and seek a u + iv that is regular in the domain B and that maps B onto
the n-sbeeted strip lying over G and having the property that w -~ +
00
i ~s
Z
approaches points ak through points in B. To construct the function w
=
f{z), let us choose on each of the curves K k ,
for k = 1,' . " n, a point b k = Zk (tk), where 0 < tk < 1. These points, together with the points ak partition the curves K k into arcs K~: z = z k (t) for 0 < t < t k F
and
K;::
z = z k (t) for tk
< t < 1.
Suppose that u (z) is a bounded harmonic func
tion in B that is equal to zero on the arcs
K~
and equal to 1 on the
a~cs K;:.
function
-1T-
I
log
(l/Iz - bkl),
tb~ n-sheeted strip 0
< lR (w) < 1.
-00
uniformly at tbe same speed as the
z
as
Let Wo denote any point in tbe strip 0
<
. lit (w) < 1. Since the function f(z) does not assume the vaiue Wo in sufficiently small neighborhoods of tbe points ak and b k for k,= 1,"', n, tbe number of zeros of tbe function f(z) _. Wo in B is equal to (217)-1 times the increment in arg [f{z) - wo] as z moves around tbe boundary K '= U~=IKk of the domain B in the positive direction, where, in ueighborboods of the points ak and bk, tbe patb is along small arcs contained in B. But as z moves around the curve K lc (for k = 1,"', n) in the positive direction, beginning at some point Zo € K k , tbe corre
Let v (z) denote the harmonic conjugate of u (z) (defined up to a constant term).
sponding point w = f{z) firSt moves downward along the imaginary axis and then
In general, the function v(z) is not single-valued in B. Let
moves along a sufficiently distant curve to the line
denote the increase in v(z) as z moves around K k
for k
1, "', n, in tbe positive direction. Wk>
=
lR (w)
'=
1 and then moves up
ward along tbat line. It then moves along a sufficiently distant curve back to tbe imaginary axis and returns along it to the initial point zoo Tbis shows that the
Here, in a neighborhood of the points ak and bk , the motion is along small arcs contained in B, The numbers W k depend on the choice of the points b k' If we
increment in arg [f{z) - wo] as z moves around the curve K lc in the positive
succeed in choosing the points b k in such a way that all the numbers
direction is equal to 21T. Then, the increment in arg [({z) - wo] as z moves
equal to zero, then the function f(z)
=
W
k
are
u + iv will be single-valued and hence
around the entire boundary of K intbe positive direction is equal to 277n. Conse
o in B is equal to n no mat
regular in B. Let us show that, in such a case, the function f(z) serves as the
quently, the number of zeros of tbe function f(z) -
required mapping. To do this, let us investigate first the behavior of f(z) in a
ter wbat the value of Wo; that is, the function w
=
neighborhood of the points ak and b k • Let us map the simply connected domain bounded only by the curve K k and containing the domain B onto the halfplane
domain B bijectively onto the n-sheeted strip 0
< lR (w) < 1.
'2'(() > 0 in such a way that the points ak and bk are mapped respectively into o and 08. The mapping function z = ef>{() is regular at -;; = 0 and has an expan sion z = ef> {()
=
ak
+
C 1 (+
.. " where
CI
-f.
0, in a neighborhood of that point.
values 0 and 1 on tbe positive and negative halves of the real axis respectively. But the function (1/1T) arg ( has this same property. Therefore, the difference
f{ef> {()) - (1/ rri) log ( can, in accordance with the symmetry principle, be continued
f{z) does indeed map the We assume that the
function f(z) is single-valued in B. Let us show tbat values t b .. " t n for wbich all the numbers WI.···. are equal to zero do exist. Choosing t n arbitrarily in the interval 0 us consider the different tk such that 0
This transformation maps u (z) into a function tbat is harmonic and bounded in a domain of tbe (-plane tbat contains the real axis on its boundary and tbat has tbe
W
Then u(z), v(z), and
Wk>
k
=
:s tk :s 1 for
k
< t n < 1,
t l " " , tn_I'
set
V(Z)=V(Z, tt> ... , tn-I),
Wk=Wk(tt> .. " tn-I),
let
= 1,2,' ", n - 1.
1,"', n, are functions of
U(Z)=ll(Z, tt> ... , tn-I),
W n
k= 1, ...• n.
We
280
§s.
VI. MAPPING ONTO A DISK
Let us show that there exist numbers t I>
•• "
Wk(tl> ... , tn-I)=O.
t,,_1
such that
that t!. v. k
I) The functions (U k (t I>
S I.
• ",
k= 1, 2, ... , n.
(1)
=
dh(!.,
Z)=2~.! (h(bk,
iLtty of the domain is equal to O. On the other hand, this increment is equal
z)-h(ak' z»,
S tv S I
4) It follows from properties 2) and 3) that (Uk(tI>"', t"_I) (for k= I,'"
••• , n - I) is a decreasing function of the parameter tk'
for
5) The function (Uk(tI>"', tn-I) is positive for tk = 0 and negative for I (under the assumption that all the other numbers t l' •• " t n _ 1 are different [from both 0 and I). To see this, note that for tk = 0, the function u is equal to (unity on K k and less than unity in B. Consequently, au/an < 0 on K k • Also, we iLtk
au = d.\ dv= d\ - au oydx+ ax dy k
It.
where the integral is over an arbitrary closed analytic curve
S tv S I,
on the basis of the formula
tv' the difference 2) Each (U k (t I>
= jdV=- } ~~dS k
•• "
11•• k=(J)k(t~, ... , t., ... , tn_I)-wk(t h
t~, ... , tn_I) is the increment in the function vv{z) as z moves around the contour :K k • This function is the conjugate of the function U v (z), which is harmonic in B and van ... ,
ishes on K except on the arc of the curve K v between the points z v (t) and
Now, by using properties 1)-5), let us show that the system (1) has exactly (one solution. It follows from properties I) and 5) that, for every system of values
t2, ... , t,,_1> there exists a unique value t 1
Uv + iv v vanishes at some point on K k and hence the image of the domain B under the function w = f)z) would get OUtside the strip 0< ~(w)
n
!connected domain by means of suitable cuts, the function v{z) will be single
tn-I)=2~.! ~ ••••
t"_I) increases with increasing
.. "
~ ..., tn-I)
t,,_I):
I, .. '. n - 1. This is true because, in
k=IKk
and hence u{z, t I>
"
3) We have
t,,_I) are continuous functions of each of the
where v n
u(z. tl> ....
and this means that (U k (t l'
tv for v ~ k.
To do this. we mention first a few properties of the functions (U k (t 1> •• variables tv for 0 S t}J accordance with §3.
< 0,
281
MAPPING OF AN n-CONNECTED OOMAIN
~ with increasing :i. "
,that
'1' it follows from the identity w! ['tl
'1
(tlb
••• ,
tn_V,
.ti ,
... ,
tn_I]
== 0
(t2," " tn_I) increases with increasing tv for v ~ 1.
Let us define the functions
(J)k (t~,
... , tn_I) = Wk ('tl (tlb
••• ,
tn_I), t 2,
... ,
tn_I), k
=
2, ... , n.
r ..~-_-- -_·_·-·_"~ __"""-"-"'--,""~-~----""__ ~---viiM'iR"iiiO"''''''''""''''''''-----'----·_-'''''''''----';;;-"-~,,,,,,~=«=>~,-,,, ""'''''T'''''''~''-''-'''''''''---~''''''"'''''"',,"'''''''''''';''ii'''''"'''''''''''''''''-'''''_~'''''''''''''''''''''''--;;'''''''''''''''''''~~''"''''''''' ......
282
VI. MAPPING ONTO A DISK
..-"...-..;;-._,,....,...-"""""..........,~""' ....................."""""
§6. SOME IDENTITIES
283
These functions are continuous with respect to t2' " ' , tn_I' For each k, the
and c on
function w~ (t2,"', tn_I) obviously increases with increasing til for
w = f(z) is uniquely determined. This completes the proof of the theorem.
II
~ k.
Furthermore, it follows from (2) that, for all t2' " ' , tn_I'
Kn are mapped respectively into
+ ioo, 8, and - ioo and the function
The proof that we have given is due to Grunsky. 1) The question arises whether the idea of the proofs of numerous existence theorems can be used in the
n
~ Wk(t2,
present case to establish the existence of functions mapping an n-connected
tn_I)=O,
"0'
k=2
domain B onto an n-sheeted disk by using some extremal property of these func
It then follows that W~ (t2,' ", tn-I)
(lor k
=
2,"', n - 1) increases with
tions. It turns out that this is possible and that in this connection the problem
If'
(a)!, where a is a finite member of B,
increasing tk' Finally, it follows from property 5) that w~ (t2, .• " tn_I) is posi
centers around the maximum value of
1. Consequendy, the functions W~(t2>'''' tn_I) 1)-5), with the number of functions and the number of variables decreased by one. We reason in the salIle ",ay with
out of all the functions f(z) that satisfy the condition f(a) = 0, that are regular
tive for tk
=
0 and negative for tk
=
satisfy conditions analogous to conditions
these functions: there exists a unique value t2
=
T2 (t3"", tn_I)fin the interval
0< t2 < 1 that satisfies the equation W~(t2,"" tn_I) = O. Here, T2 (t3"", tn_I) is continuous and increases with increasing til for II > 2. Now let us consider the functions
Wk (ta, ... , tn_I)
= wi. (t2 (ta, .0.,
tn_I), t a, .•• , tn_I), k
= 3,
0", n,
and continue in the same way. After n - 1 steps, we arrive at the uniquely de
such that Wok
tn_I
(Tb ••• ,
=
canst, t n_9= t n _2 (tn_I), ... , t l = "I (t 2 , • tn_I), Tn_I) = 0 for k = 1, .. " n - 1. But then, in accordance with 0
a function w = f(z) mapping the domain B bijectively onto the n-sheeted strip in such a way that the given points ak €
mapped into w
=
will be explained in Chapters IX and X.
§6. Some identities connecting a univalent conformal mapping and the Dirichlet problem In §4 of Chapter V, by using the method of contour integration, we estab lished certain relationships connecting various functions that map a given finitely we shall continue the application of this method 2 ) by introducing into our study
.,
equation (2), we also have w n (Tl> •• " Tn_I) = O. This proves that there exists a system of values t 1> •• " tn_1 satisfying equations (1) and hence that there exists
o < ~w < 1
Chapter XI. It uses a number of boundary properties of analytic functions which
connected domain univalently onto canonical domains. In the present section,
fined system of values
tn_I =
'in B, and that do not exceed unity in absolute value. The proof is given in §3 of
K k , k = 1, 2, •. "
n, are
+ ioe.
appropriate harmonic functions defined as the solutions of certain Dirichlet pro blems. Specifically, suppose that B is an n-connected domain that does not in clude "" and that is bounded by n closed analytic Jordan curves KII for II = 1,,·· ... , n. Suppose that w)z) (for II = 1, ... , n) is a function that is harmonic in
B, equal to 1 on K II , and equal to zero on all KIJ.' for 11 ~ functions wlI(z), for
for k
N (15)
k=1
~ (~rkk. (ak) - A:)=o, 'J= 1, ... , n. k=l
(e
An
~ "" i9w.(ak»=O, ' v=l, ... ,·n, L..Jrk..J(e =
1, " ' , n. Consequently, condi
tions (17) are satisfied for the numbers 'kei6 and ak, for k = 1,,", n. There fore, there exists a constant A such that the corresponding function feu) defined
in accordance with formula (14), which, in the present case, is the function
Since fez) is real on KI)' we have N
11o
= ••• =
n
=2 hffi(rkQ(Il, ak»- hrkk.Jak)+A. k=1
Consequently, the rank of the matrix
i6
(16)
n
n
k=1
k=1
f(z)= ~rk(eiOQ(z, ak)-e- i9 p(z, ak»+A= ~rkeIOj_O"Q(z,
n'
In the first place, it follows from formula (1) of §2 of Chapter V that, for z,
F~Z)= !(rkQ(Z, ak)-rkP(z, ak»+A, 1 "=1
p; (z,
We confine ourselves to proving this assertion only for the first of the func
them, keeping (19) in mind, we obtain the following formulas:
I
/9 (z) =
tions (23), assuming, as above, that the domain B is bounded by closed analytic
are obviously Schottky functions in the domain B. If we apply formula (14) to
F(z)+
..\'
(~
I(I < 1.
Jordan curves K.",. v and
u)
n'
where u, v E: B,- is such a function and each of them maps the domain B onto the
=
Then, the functions
F(z)+F~Z)
P; (z,
·- ;·:·:~-:· :~-~-
l
"'.0.
,,;,
~
t-
In the second place, if z = zv(s) is the equation of the curve K v and s is the length of arc on K v measured in the positive direction with respect to the domain B, t hen, when we differentiate (4) with respect to s, we obtain
r
292
\
VI. MAPPING ONTO A DISK
P'z (z. (S), u) z~ (z) = It follows that on KJ)'
(z, 1ft (z) 1= p; Q; (z,
I
Q; (z. (S),
ll) Z~ (S).
(25)
I-I 1=1.
(8) Z' (8)
u) u) -
Z'
(26)
On this basis, we conclude that the functions P; (Z, u) and no zeros on K. To see this, suppose, for ex.ample, that a €
Q; (a,
K. Then, in accordance with (26), we have P; (a, u)
j; 12 (a, u) = O.
have j ~ (a, u)
u)
Q; (z, =
U) have
THE MEASURE PROPERTIES OF CLOSED SETS IN THE PLANE
0, where
0 .and hence j~ (a, u)
=
CHAPTER VII
=
Therefore, in accordance with formula (1) of §2 of Chapter V, we
=0 for all e. This means that, for all e. the mapping ,= j iz, u)
maps the point a into the endpoint of the boundary cut of the image of the domain
B. On the other hand, if in a neighborhood of z j~ (z, u)
= Cl (z -
a)
=
a we set
+ ..., I "2" (z, u) =
d l (z - a)
The transfinite diameter and ~eby~ev's constant
In contemporary questions in the theory of functions of a complex variable, a
significant role is played by certain specific ways of measuring closed sets in
+ ...,
the complex plane. We begin with one of these ways, which was proposed by
we conclude from the geometrical interpretation of the argument of the derivative of an analytic function that the ratio c 1/d 1 is purely imaginary. Therefore, since
je (z, u) = eiO (Ct cos 9 -ld t sin 9) (z - a)
•
§I.
+ ...,
Fekete. 1) Let E denote a closed bounded infinite set of points in the z-plane. For n points z l' Z 2' .• "
e,
(arg
Q; (z,
P; (z, u»
K =
u»
K = -
B,
V(z l'
417 (n - 1).
z) is equal to the Vandermonde determinant
Z l' .. , , Z •
"
"
z) is a continuous function of its arguments and the fact that the set
E is closed. Finally, let us define which is regular in the
=
exactly 2n zeros in B. This information, Cauchy's theorem on the number of zeros, and property (25) enable us to conclude that the function ~ =
ft (z)
assumes
< 1 at exactly 2n points and it does
(2)
does not increase with increasing n. Let ';1' ';2' ... " ' , ';n+l denote a system of points belonging to the set E such that
Let us show that d
IV(';I' ';2' "', ';n+l)!
not assume any other values at all; that is, it maps the domain B onto a 2n sheeted disk. This completes the proof of the assertion. 1 )
2
d V'! (n-l) , nil'
has exactly 2n - 2 zeros in B. But then the function fl (z) has
I"
(1)
. longing to the set E. Such a maximum exists by virtue of the fact that
477, it follows from what we have proved thac
in the domain B every value in the disk
n ~ 2.
n Let Vn = Vn (E) denote the maximum value of IV( z l' " ' , z)1 as z l' " ' , Z n range over all systems of n distinct points be
(arg Z'Z' (8») (8) K= 2 (arg z' (S»K= 41t (n - 2).
P; (z, u),
Zl)'
k,l=l k
... ...
I
n-I Xn+1 t n (Xn+t)
~ (n
+1) mil V
m i.e.
n
-0::
Vn +1 Vn
,,;::::
-=:::
(n
IV(X 1'
0/2
of
m> 0 such that Ian -
al
m. Then, for n > m,
+ as +n ... + an -a I I.~ (a. =_1
m
a)
~ (a. -a)
,;:;; I _.=---=--1
+n-m~
_
n
n
2'
d2. d2.
This proves
the convergence asserted in the lemma. The above reasoning remains valid when we are considering polynomials with zeros only on E. This means that if we set ;~ ~ n that r ----> r = d as n - ..... 00.
" ' , Xn+ 1)1 = Vn+ l'
=
1;;; , we also obtain the result n
n
Summarizing what has been said, we have
VV;l ~ (n + 1) m n•
+1)
denote a positive number. Choose
But, for sufficiently large n, the first term is also less than
Theorem 1. For an arbitrary closed bounded set E of points in the complex
plane, the transfinite diameter d and the tJebysev constants
Thus, inequalities (8) are proved. They can be written in the form 't n ,,;::::.
f
The second term in the expression on the right is less .than
, Xn+l) I
Now, if Xl' " ' , X 1 are points on E such that n+ we obtain
The arithmetic mean of the first n (n +
n
+ + I tn (Xn+l) II VeX!> ... , xn)l.
Vn +t
for n
XII+l
n-l
T.
follows from
Proof. Let
d2
1
If we expand the determinant in terms of elements of the last row, we obtain
+1
T
ai
n --I XI in (Xt)
to log
Lemma 1. If a sequence of real numbers al' a 2, ••• converges to a finite
ceed Vn+ II V. n For x E E, the modulus of the left-hand member is no less than -
1)
limit a, the n the sequence of their arithmetic means (a l + a 2 + .•• + a)1 n also converges to a as n _ ""'.
Vn, we 'conclude that the modulus of the right-hand member does not ex'
m n • ·Consequently, mn .
297
these logarithms is the logarithm of the left-hand member of (9). That it converges
.. e) •
itt ... ,
v
TRANSFINITE DIAMETER AND CEBYSEV'S CONSTANT
as n ----> 00. Since Tn ---+ T as n - ""', the sequence T l' T 2' T 2' T 3' T 3' T 3' " ' , in which T·n is repeated n times, also converges to T and the sequence of their
(8)
In the first place, we have (x
v
§l.
VII. MEASURE PROPERTIES OF CLOSED SETS
T
and ;- coincide:
d = T = ;-.
't n
II'
From this theorem follows the simple fact that, if the entire boundary of some bounded domain belongs to the set E, then the transfinite diameter does not
Now for n let us set the successive numbers 2, 3, .. " n and multiply the reo sulting inequalities. We obtain 2
change if we adjoin to E the interior points of that domain. Resting on the identity of the transfinite diameter and Cebysev's constant,
2
('t; • ~. 't~)n (n+ I). (v~)n (n+ I) ~
~
we shall give two simple examples of the calculation of the transfinite diameter.
dn+1
~ [en
+ 1) !]n(n
2 + I)
Bu' (V ,)2/0'0,1) 1 uud [(n + 1),]2/0'0'" _ show that d = T it remains only to show that
('t~
2
...
1. Suppose that E is the disk
2
't~)II(II+I) (v~)n(n+ I). /
1 as n
~~. C~Y'
1zl < R.
Since
2",
1 "2 $
,u
~
\Pn (z) I~ dB =R~Ii+ ...
+Ien I~ ~ R~II,
2
2-( 't 1't2 ... 't~n(n+I)_'t
(9)
1)
00,
we have
V d(E)
::; d(E*). Therefore, equation (10) must hold. This
(that is, m::; n) because inside each of them there is at least one root of the poly
nomial p (z). This last fact follows from the fact that otherwise we could conclude,
completes the proof of the theorem.
on the basis of the maximum principle as applied to
If we apply Theorem 2 to the case in which E is a disk, we obtain the Corollary. The set of points z defined by the inequality Ip(z)! ::; R, where p (z) = zn
+ ... + C n has transfinite diameter d =
n.
= const.
:,'of the Kk' that p(z)
,K
the point
,
k
(=
1(1 = pl/m k•
f(~mk)
of this nature, dealing with outer Jordan 'content. Let us review the definition of
E denote a bounded set in the z·plane. Let us cover the plane
with a grid of squares of side lin and let us denote by s n the sum of the areas of those closed squares that contain at least one point belonging to E. The limit as s
---->
00
Therefore, on
1(\
=
00
set E does not exceed TTd 2, where d is the transfinite diameter of E. Proof. Let us first prove the theorem for the special case when the set E is
VR.
(k),m k '0
(2)
•
Equation (2), with (= pe' 0, if n is sufficiently great we have It n (z)1 :s (r + () n on E. Consequently, the set E lies entirely in the interior of finite domains bounded by the curves de
any
(r + ()n. Therefore, the two-dimensional outer Jor dan content of the set E does not exceed the sum of the areas of these domains and by virtue of what was shown above, this sum in turn does not exceed 17(r + ().2. / Letting (---> 0, we obtain the assertion of the theorem. This completes the proof of the theorem. fined by the equation It n (z)1
=
Theorem 2. The one-dimensional outer Jordan content of the orthogonal pro jection of a closed bounded set E onto an arbitrary straight line does not exceed 4d, where d is the transfinite diameter of E. Proof. Obviously, it will be sufficient to prove the theorem for the projection onto the real axis. Again, we shall consider first the case in which E is the set defined by the inequality
Ip(z)l:s R,
where p(z) = zn + ..• + c ' n
Suppose that
the real axis, it must be a multiply connected domain. Since IP (z)! = R on the exterior boundary curve of this domain, it follows on the basis of the maximum principle that IP (z)1 is less than rather equal to R on the interior boundary curves. Furthermore, each of these domains contains one of the segments because the interior of anyone of them that did not would not contain roots of the polyno mial P (z) (since all the roots of P (z) are real), so that, again by virtue of the maximum principle, we conclude that P (z)= const. Now, if there were no roots of
P (z) on one of these segments, there would also be none anywhere in the entire domain containing' it and, consequently, we would again conclude that P (z) == const. This proves the assertion made. Let I denote the farthest right of all the segments containing P and suppose that I contains m roots of the polynomial P (z) (without consideration of their multiplicity). We have 1:S m
:s n
with m = n only in the case in which P consists of a
single segment. In addition to P (x), let us construct polynomials PI (x),
P 2(x), ., ·of degree n with real coefficients and in particular with the coefficient of x n equal to 1 in all cases.
By P k' I k' and m k' we mean the same thing for Pk(x) as we mean by P, I, and m for P (x). Let us show that, if m < n, there ex . ists a polynomial PI (x) such that m 1 > m and the sum of the lengths of the seg ments constituting PI is greater than the corresponding sum for P. To see this, suppose that P(x) = Q(x)R(x), where Q(x) is a polynomial with zeros on I and R (x) is a polynomial with zeros outside I. Furthermore, let d denote the dis tance from I to P -I. Let us define PI (x)
n
p(z)=
domains contains at least two segments. Then, by virtue of its symm/try about
n (Z-ik),
"=1
Z=x+iy,
i"=ak+ib,,.
IQ(x + d)!
< IQ(x)1
=
Q(x + J)R(x). If x € P -I then
because each linear factor in the expression for Q(x + d) is
less in absolute value than the corresponding factor in Q(x) by an amount d. Consequently, if x € P - I, then
IP 1 (x) I < IP (x)1 :s R.
This shows that
=,=:",,,,,=====,~,,~=:"";''------~=--~-~=--'-=-----'-'-:---':;;;:;';~;;;o7::':;;:--'-~---::--.---
304
--.-:-;;:;:.;;--;;;--_-;;;..'?----..- -.. -;;.---;~--.-.-:__~;-:;.~~.~-
._;;;:;~"._
- ;....::.-
"'--.::'~,-.~.-~'""='''',,;:-_-:::.::__.5:::::;,~·;;.,:_:.-
.._ "".".. ,.,.,~c'".~"._
", _"',:..","_"_'':;''''''''':~::~''''''''''~,~'''''~--:'"''-'
;'~;~",.;;,:,~-;,;::;;:~-;~
--
P -I €
Pl' On the other hand, if x E I, we have analogously IP 1 (x - d)1 = d)! < I Q(x)R(x)! = IP(x)1 < R and, consequently, the inlerval [' ob tained from I by a displacement to the left by an amount d belongs to Pl' Thus,
I(z -
~) I.:;;; 1,
Cl) (z -
IQ(x)R(x-
PI contains P -I and ['.Consequently, the sum of the lengths of the segments composing it is not less than the corresponding sum in the case of P. Since P (x) 1 has all its zeros on P - I.and on 1', the number of segments constituting P 1 is less by at least unity than the number of segments constituting P. Reasoning
k::;
Cl> 1.
Obviously, the transfinite diameter of this set is equal to 1. The boundary of the domain B complementary to E and containing
I(z -
00
is defined by the equation
a) (z - 1/ a)! = 1, that is, by the equation
- !(z -l+exl)1 2ex
(ex-l - l 2ex
)11_ -1
n - 1 of steps at a polyno
> 2a,
mial P k (x) such that P k consists of a single segment and has length no less
and, for a 2 - 1
than the sum of the lengths of the segments constituting P. But if the segment
boundary of the domain B lie the roots of the equation (z - a) (z - l/a)
P k has length L, then, by §1, its transfinite diameter is equal to L/ 4. On the other hand, P is contained in the set of points defined by the inequality IPk(z)1 ::;
R and the transfinite diameter of that set is, in accordance with §1, equal n,-
n~
to V R. Consequently, L/4 ::;~~-~--matls, L::; 4
n
VR
=
4d, which proves the
"
n
v
the Lebysev polynomial and rthe Cebysev constant for the set E. Then, for given
(> 0
and sufficiently great n, we have Itn(z)
I::; (r + d n
1, that is, the points 0 and a + 1/ a. The difference between these two roots, which is , equal to a + 1/ a, can be made arbitrarily great bya suitable choice of a. 'In this case, the ordinary diameter of the set E can also be made arbitrarily great. =
We mention some consequences of Theorems 1 and 2 for a set E with zero by virtue of Theorem 1, equal to zero. It remains equal to 0 in the case in which
Now, let us turn to the general case of the theorem. Suppose that t (z) is v
consists of two ovals without points in common. On the
transfinite diameter. In this case, the two-dimensional Jordan content for E is,
theorem as regards the special case in question. )(
305
§2. BOUNDS FOR THE TRANSFINITE DIAMETER
VII. MEASURE PROPERTIES OF CLOSED SETS
analogously with PI (x), we arrive after some number
~c-~ ... _" ..,.,,,.. ';"'';;'-''',,"'. ""-",-,:.,,-~,".~:;;:;--:,c;;:;;:;.;:.~,,,~-~_,;;:,"~"_:;;;:_:;;::;;::;:,,,::~:::-,~ -.-£-~-",,::. "":,,~~£;:::'"'- ~"';':;;'::C?,~;;~-;;_';;':';~'
on E. But, by virtue of
all complementary domains not containing
00
are adjoined to E because this does
not change the transfinite diameter . .It follows that E has no complementary do Thus, if dee) = 0, the complement to the set E is the
what was said above, the projection of the set of points defined by this inequality
mains not containing
onto an arbitrary straight line does not exceed 4 (r + d.Consequendy, the projec
unique domain containing
tion of the set E onto the same straight line has one-dimensional outer Jordan
a boundary point of that domain. Theorem 2 proves more than this, namely, that
content not exceeding 4 (r + (), that is, by virtue of the arbitrariness of
(> 0,
not
exceeding 4r. This completes the proof of the theorem.
If the set E is a continuum, we easily obtain from Theorem 2 the result that the ordinary diameter of the set E does not exceed 4d. As we shall see in §3, this result is equivalent to a bound on the diameter of the boundary of the image of the domain
1'1 > R
under a univalent mapping by the function z
= ,
+
ao +
00.
00
and every point of the plane is either an interior or
E does not contain any continuum because if it did there would be a straight line such that the projection of the set E onto it would have a positive one-dimensional outer Jordan content. These are necessary conditions for d (E) to be equal to O. They are not, however, sufficient. As Nevanlinna
1)
has shown, there exist sets
that do not contain a continuum but that have a positive transfinite diameter. We shall not stop here to construct such sets.
a / ' + . " and it follows directly from §4 of Chapter II. Thus, Theorem 2 is a
In' conclusion, we present another interesting theorem.
generalization of this property of simply connected domains to multiply connected
Theorem 3.
domains if by the sets E we mean the boundaries of domains. The question arises, is it possible in the case of an arbitrary closed bounded set to conclude on the basis of its transfinite diameter whether there exists an upper bound for its ordi
1)
Let E denote a closed bounded set in the z-plane and let B
denote the domain complementary to E that contains 00. If fez) is regular in B and ifit has the expansion fez) = a z- 1 + a 2 z- 2 + .•. in a neighborhood of 00, 1 then, by setting
nary diameter, as is the case for a continuum? The following example shows that the answer to this question is negative. Let E denote the closed bounded set of points defined by the inequality
_/
1) 0, there exists a closed
with boundary consisting of a finite number of closed analytic Jordan curves. We denote this boundary by K. Suppose that E* has transfinite diameter d*
where L is the length of K. Therefore,
. lim n.;t, An I.~ d
(Zk -
ZZ)
k,Z=1 k N, I { 16 - e
)n :::;; AAn_ n
1
(1
The importance of the constants d and
)n
~ 16+ e .
ities obtained, we get
2
p(2N+p+1l
1
1
__ 1_
A~'HP)2 :::;;AAr-+1pP)2 ~(16+e)
Let us again look at a closed bounded set E of points in the z-plane. The complement to this set in the z-plane consists of a finite or countable set of dis us show that one can exhaust this domain by means of domains B(n), where n=1,
2
2, ' , .; each ·of which is bounded by a finite number of closed Jordan curves. Here, exhaustion is understood in the sense that [j(n)C B(n+l), B(n)C B, and every
or
16- e) 2(N+p)2
defined in the preceding section
joint domains. We denote by B that one of these domains that includes 00. Let
AN + p (116+ e)P(2N~P+1l 1 e)P(2N+P+I) ~-A--~ (16N
(
T
follow from their association with the Green's function.
If we now give to n the values N + 1, N + 2, •• " N + p and multiply the inequal
1
To
gers, it follows that (6) with d < 1 can hold only under the condition that tbese
[
This last limit can be calculated with the aid of Stirling's formula. We conclude from (10) that, for arbitrary positive
< 1, then it is a rational function.
see this, note that since, in the present case, all the determinants (5) are inte
From this formula we get
r
309
the theorem of P6lya: If two coefficients in the expansion fez) = at z-t +
rows and columns, we finally obtain the recursion formula
[
;;:';;;;;.,u;;::;;::-...."~_;;;;;:~;;;:';;,;-;,,:,.,,.:,..,;i:t-~t'i1;.':,:;.:;;;;.::;:~.~,~_._--:;':';;:;..:;;
§3. CAPACITY OF A CLOSED BOUNDED SE,l
VII. MEASURE PROPERTIES OF CLOSED SETS
A -
-
p(2N+p+1l 2lN+p)2
_ _I
A~N+P)2
point z € B beginning with some n lies in B(n). Let us cover the domain B with a grid of squa~es of side
El2
and let us consider the set consisting of points of
the closed squares in the grid that contain at least one point belonging to E. The By virtue of the arbitrariness of
f
> 0, when we let p approach 00, we get
complement of this set consists of domains one of which is the domain B con f
n2
-
1
taining 00. This domain, together with its boundary, lies in B. The distance from
lim yA n =4'
its boundary points to the set E does not exceed
n-+oo
f.
By extending the domain B, f
Consequendy, in accordance with Theorem 3, we have d ~ 1/4. This is.a sharp
we can arrange for its boundary to consist of disjoint closed Jordan curves. To
bound for d because, if E is the segment connecting the points
do this, we need only adjoin to B
°
and 1, we
have d = 1/4 in accordance with §L As will be pointed out in §5 of Cbapter XI, inequality (6) remains a sharp inequality for an arbitrary set E consisting of a finite number of disjoint closed analytic Jordan curves. replace d in (6) with an arbitrary number d'
< d,
1)
This means that, if we
this inequality will fail to hold
for some function fez) that is regular in the domain B. In that section, we shall
f
sufficiently small neighborhoods of the mul
tiple points of the boundary of Bf • By giving
°
f
suitable values
f
n
such that
f n as n -.. 00, we obtain a sequence of domains B(n), where n = 1, 2, .•• , that exhaust B. One can easily show that if the domains B(n) exhaust B, then,
for an arbitrary given
f> 0, when n is sufficiently great the distance between all
points of the boundary of the domain B(n) to the set E is less than
f.
Suppose now that the domain B defined above is exhausted by domains B(n)
present a theorem on the distribution of values of polynomials in the complex plane expressed in terms of the transfinite diameter. Furthermore, we mention
with boundaries K(n). The Green's function g (z, 00) for the domain B(n) is, as
briefly an actual application of Theorem 3 to power series. Specifically, we have
we know, a harmonic function in B(n) except at the point z
n
in B (n), it assumes the value
z l)Goluzin [1946b),
=
lXJ
°
= 00,
it is continuous
on K(n), and it behaves in a neighborhood of
in such a way that the limit
310
S3.
VII. MEASURE PROPERTIES OF CLOSED SETS
it is necessary and sufficient that the capacity of its boundary be positive:
exists and is finite. The quantity y n is called Robin's constant for the domain Consequently, in a neighborhood of z
=
00, we have
the plane is equal to its transfinite diameter; that is C(E)
where u n (z) is a harmonic function in B(n) including the point z = 00, and ap • _ . proaches 0 as z ----> 00. Since B(n) C B(n+1), it follows that the function n+
l(z, 00) - g (z, 00) is harmonic in n
negative on the boundary
including the point 00, and it is non-
B(n),
In accordance with the maximum principle, this
K(n).
last statement also holds everywhere in the domain
B(n);
With regard to capacity, we shall now prove Theorem 2.1) The capacity of an arbitrary closed bounded set E of points in
g" (z, 00) = log I z I+i" +u" (z),
g
•
Theorem 1. For a domain B containing 00 to have a Green's function g(z,oo),
lim (g,,(z, oo)-loglz\)=i" 18-+(;0
B(n).
that is, for all z E
Green's function and that h(z, 00) is the conjugate function to g(z, (). For arbi trary finite z E B, the function II
(~, z) = g(~,
z) - g(~, 00)+ log I~ - zl,
as a function of (, is harmonic in B. If we apply formula (5) of to it, keeping in mind the fact that u«(, z) = log
Yn + U n (z) either converges everywhere to + 00 in B or converges to a harmonic function y + u(z) such that u(oo) = O. In the latter case, the quantity y is called
(, z E B, II
Robin's constant for the domain B, the quantity C = e - Y is called the capacity =
log Iz 1+ y
+ u(z)
is called the Green's
function of the domain B. Obviously, C and G(z, 00) are independent of the B(n).
Green's function g(z, 00) does not assume
B, all its limiting values are nonnegative. However, we cannot assert that these limiting values are always equal to zero. For example, at isolated boundary points of the domain B, the value of g (z, 00) will indeed be positive because, in a neighborhood of any point, the function g( z, 00) is a harmonic function thatis'/ bounded below, so that it is also harmonic at that point itself. We note that, in the case in which y n + u n (z) -
00 for z in B, it is also true that y n ----> 00 be cause u n (00) = O. . In this case, the capacity of the set E is taken equal to zero.
g(z, oo)-i=
B(n)
that exhaust the domain
tively by the inequalities functions for the domains
g(z, 00) B(n)
> i n,
B
let us take domains defined respec
where
i
n
---->
0 as n
---->
g(z, 00) - y, where y is Robin's
2~ ~ logl~'-zldh(~', 00).
(1)
T
= e- Y , where
T
is C=ebyS'ev's con
stant for K. Let A denote a positive number sufficiently small that the inequality g(z, 00) > A defines in B a domain BA with boundary K A consisting of closed Jordan curves that approximate the corresponding curves constituting parts of K.
Since
2~ l'.~
dh (C'. 00) = 1
K
'
let us partition the boundary K by points (", for k =1, . , " n (numbered in order of occurrence as one moves around
n
they converge in B to g(z, 00). The result that we obtained above can be formulated somewhat differently as
=
Making use of this formula, let us show that
00. The Green's i , and
are respectively the functions g(z, 00) -
(~', Q.
K
finite number of closed Jordan curves, with the Green's function defined in Chap· the domains
z Idh
zion K, we obtain, for
constant for the d,omain B, we have
have given here coincides, when the boundary of the domain B consists of a
is a Green's function in the previous sense, for
-2~ .~ log I~' -
If we now set (= 00 and note that u(oo, z)
It should be noted that the concept of the Green's function g(z, 00) that we
Specifically, if g(z, 00)
(C, z) =
I( -
S3 of Chapter VI
K
negative values anywhere in B. Consequently, as we let z approach the boundary
ter VI.
deE).
of a finite number of disjoint closed Jordan curves. Suppose that g(z, () is its
It follows in accordance with Harnack's theorem that the sequence of the functions
of the set E, and the function g(z, 00)
=
B containing 00 with boundary K consisting
Proof. Consider first a domain
B(n),
gn+1 (z, 00) ~g" (z, 00), in+l ~i".
choice of exhausting domains
311
CAPACITY OF A CLOSED BOUNDED SET
1) Szego [1924].
K, into subsets 1" such that
....
, __ ._.....,-,
-T_~_~~'
312
---------- ------ -
---_.,
;.••-'-
-
-".~,,-~
-iii...';;'>;-;>';""
;'~k.",,,,"-;""",,,,,.~,,_,,,,,,,,·,=_-,,
VII. MEASURE PROPERTIES OF CLOSED SETS
2~ ~
dh (C', co)
= ~,
...,,,,;_ _.
§3.
·,"'''''''"";.=.,='''·>:'%''''''''"'',,'''''''''''''''·'''''''.. ., O.
The question arises as to ways of telling when the harmonic measure of a given set a is zero or positive. The vanishing or not of a harmonic measure
It follows from this last assertion that, if the harmonic measure of the set a
w (z, a, Bf3) holds simultaneously for all points of a domain B (3 because, if it vanishes at a single point in B 13' it must vanish everywhere in Bf3 by virtue of
with respect to some domain complementary to a is equal to zero, then a does not contain a continuum and the complement to a is the unique domain including
the maximum principle. However, the same phenomenon holds with regard to the
00.
We now point out the following simple properties of sets of zero harmonic
curves {3 chosen in B. To see this, suppose that for some curve {3, we have
measure.
{3'" denote another curve of the same ki~d. Let y and y' denote closed Jordan curves lying in B and enclosing {3 and {3 , such that y' lies inside y and there are no points of the set a between y and {3 *• The curves y and (3* together form the boundary of a domain B 0 contained in B. Set () = maxz€y' w(z, y, B )' 0 < () < 1. Now, let l denote any positive number. :For o n sufficiently large and for z E y, we have w(z, an' B~») < f because the con vergence of the sequence {w(z, an' B~»)\ to zero in B 13 implies its uniform con vergence to 0 inside B13' Let us denote by A and A' the maxima of the function w (z, a n' B~:) on y and y'. Since the difference w(z, an' B~~») - w(z, an' B~») is Ii harmonic func tion in the portion B~n) of the domain B(n) that is bounded by the curves an and y' and since it is equal to 0 on an and does not exceed A' on y', it follows that A < A' + f. Furthermore, since the difference w (z, an' B0 for
( z, all' B (II») ~ -
(J)
(Z, at. III B(II') 1
(J)
(z, tx..!, II' B(1I1) \I ,
any domain B complementary to a and for any curve {3 C B. Just as in the proof
which is harmonic in B~) is equal to zero on {3 and does not exceed zero on an'
of Riemann's theorem in §2 of Chapter II, we can, by a suitable transformation,
Consequently, in B(3 we have
arrange for the domain B to have exterior points•. Let B 0 denote the domain (J)
bounded by the curve (3 and some other closed Jordan curve lying outside B. Then, B l
(1) If a has positive capacity, then the constant y = limz-+oo(g(z, oo)-loglzl)
(4)
But
curves contained in I z I < 1 and satisfying the condition li' C B. Let us denote by g(z, 00) and g' (z, 00) the Green's functions for Band B'. Let us denote by W
p)M p'
For a point z 2 on I zl = 1 at which g' attains a maximum, which we denote by Ml' inequality (4) yields
tion without affecting the property to be proved. Let us denote this set by a and let us denote by B the domain complementary to a that includes 00. Let B' de
W
.
ffim (lJp ~ (lJp (ZII,
W
p)m p
zero.
(lJp (Zl' a', B;» mp'
:5 lim z -+ 00 (g' -
log Iz I)
'
~( I
'- _,
I
"
•
If we fix p > 1 and let B' (a sequence of domains exhausting B) approach B, then y' approaches
which we denote by m l' then inequality (2) yields
But m 1 = min I z 1=1 (g' - log I z I)
.
B p) ~ 1 - ,
(2)
B~. If zl is a point on the circle Izi = 1 at which g'(z, 00) has a minimum,
ml;;;: (1 -
, (l,
Iz I= I
(3)
W
00
and consequently min1zl=Iwp(z, a', B~) approaches
Since the function
W
p (z,
a', Bp'). on
I zl =
1 then converges uniformly to
(z, a, B p) where B p is the portion of the domain B lying in the disk I zl
< P,
it follows that QJ(z, a, B'/3) == 0 because the function wo(z, a', B'p) would =
y' and, by (1),
otherwise have a positive bound on
Iz I =
1. Therefore, the harmonic measure of
••_ -
----.- - - - - - - - .- . - - - - - - . . . - -
------
_.
,-- '-e:iiE2E".!!!
--~_.
,.'"
"..- ... _.~~'-'-"-'----_.,,'.',-
VII. MEASURE PROPERTIES OF CLOSED SETS
320
,,~,~._~
--
,-",-+
-'-""'-"--
.,,,--~~
,-_. __ .~.,-
~_.".,,-_.~,
321
§5. APPLICATION TO MEROMORPHIC FUNCTIONS
a is zero. This compl~tes the proof of the theorem.
Proof. Suppose that u (z) ~ M in B. Let us suppose that tbere exists a point
> m. It follows from the hypotheses of the theorem that there exists a point zJ € B such that u(z) < u(zo)' Let us describe a small disk (3 around z 1 in which u (z) < m l' where m < m 1 < u (z 0)' Ftom the hypothesis of z
As simple applications of the theory of harmonic measure, we shall no.. prove two theorems, one of which deals with removable singularities of harmonic func tions and the other with the maximum principle for harmonic functions.
o
in B at which u (z)
the theorem, tbe set a can be contained in tbe interior of a finite number of Jordan
Theorem 2. Let a denote a closed bounded set of points in the z-plane that is contained in a domaiT/: B. If the harmonic measure of a is zero, then every
curves lying in the complementary domain containing B. Let us denote the union of these
function u(z) that is harmonic and bounded in B except at points of the set a is
Jordan curves by an' so that, for the domain B~) with boundary an
harmonic in the entire domain B if we define it in a suitable way on a. On the
w(zo' an'
Uf3,
we bave
B';») m l' This completes tbe proof of the theorem.
An application to meromorphic functions of bounded fonn
A function f(z) is called a meromorphic function of bounded form in
This function is harmonic in B ~), equal to zero on (3, and nonnegative on an'
Izl < 1
if it can be represented in that disk as the ratio of two bounded regular functions
Consequently, v(z) + 2Mw(z, an' B~») ~ 0 in B';). Therefore, v(zo) ~ _ 2Mw > - 2Mc. Since c is an arbitrary positive number, v(zo) 2. O. Treating the
If'(z)
(1)
f(z)= '\J(z)'
function v(z) - 2Mw(z, an' B~»), we can show that v(zo) ~ O. Consequently, v(zo) = 0; that is, uo(z) = u(z) in B - a. It follows that if we define u(z) on a
i
We may assume that
I¢ (z)1
~ 1 and
11/1 (z)1
~ 1 in
Iz I < 1.
equal to uo(z), then u (z). becomes barmonic in B. On the other hand, if ·a has
A test of whether f(z) is a meromorphic function of bounded form is given by
positive harmonic measure, then, for example, the domain of definition of
Theorem 1.
Iz I < 1
w (z, a, B), which is harmonic and bounded in B - a cannot be extended to a in such a way as to make w (z, a, B) a harmonic function in B because this func
1)
Let f(z) denote a function that is meromorphic in the disk
such that f(O) f-
00.
tion, since it is equal to zero on (3, would be identically equal to zero in B - a
T(r)
and then the set a would be of zero harmonic measure. Theorem 3. Let u (z) denote a function that is harmonic and bounded above in a domain B. If
lim.;~ z. u(z) $ m for all points Z· on the boundary except for
a set of harmonic measure zero, then u(z) $ m everywhere in B.
For f(z) to be a function of bounded form, it is neces
sary and sufficient that the quantity
= d~
·i'·'...· .... , .'.
II""
Ii
~
I) R. Nevanlinna [1922].
\~ log+ If (re
2"
tB
)
I de +
I Ibkl 1, Theorem 1 will remain
Ix,
valid without the assumption that f(O) Ie- 00. One can show this by modifying the
---- ...---r
proof given above.
dh (c, 00) r
[ill \ log I/(re
=
Suppose that a function fez) is meromorphic in the disk and that it does not assume in that disk a set of values w that is a closed bounded set E of positive capacity. Then, fez) is of bounded form. 1)
Izi < 1
logl;kl'
2IT. If we now apply formula (1) of §3, we obtain
2~
2"
\ gr(/(re
~
i8
),
!
co) dB +
set K is contained in the disk jw I < R. Then, for I; €
Iwi>
r
2R, we have
I l~
log w w C
We may assume that f(O) -f- "" because in the opposite
(8)
IOg';kl'
I bkl 1 denote a number such that, for all r in the interval 0
Proof. It will be sufficient to prove the theorem for a function fez) that is
Iz I < 1.
!
)-e \dhr(e, co)]dt+
Ibkl Zg, Z,
I
=2
log 1
-1
1-21ZI ZI
--::,ZI
1-
O~usly, if z 1 ~2.'
D(Zh
z2)=D(Z2, ZI) and D(Zh
Z2)~0.
Z2
I
(9)
Z12.
Also, D(ZI' z2)=0 only
Furthermore, for arbitrary points z h z 2, and z 3 in the disk
~--
Let z 1 and z 2 denote two points of the disk
and let z 3 and z 4 denote the points of intersection of the circle
I z I=
D (Zl' Zg)
1
+ D (Zg,
of the
I
21-_2 i
1+.
1,
and arg (= arg a if a J, O.
I z \ < 1.
Iz I < 1
D (Zl' Zg) =
We can give Theorem 2 a geometrical interpretation. To do this, we define a
Iz I < 1
331
eitp . reitp+e itp l-r z,)= -el
z 2, z 3 Z4
into points (1) (2' (3 and (4, we have
(C a,
CI>
l:.J, CJ = (za,
Zb Zg,
ZI
< r < I,
and it maps the circle orthogonal to the circle
Iz I =
1 and passing
at
I 2. I+ I 2 1 I + I 21 II 2. I
~1
Z 2'
Loba~evskir, but point out that the hyperbolic distance is invariant under frac
tional-linear mappings of the disk
Iz I < I
into itself in such a way that the points icP and Z2 are mapped into the points (1=0 and (2=(Z-ZI)/(l-zl z 2)=re ,
where 0
maps the disk
I~
z 1()
metry that gives a Euclidean interpretation to the non-Euclidean geometry of
trary z 1 and Z2 in the disk \ Z I < 1. To see this, note that the function (=
z 1z)
0 and that, by
We shall not stop to construct a plane geometry based on this metric, a geo
Furthermore, the anharmonic ratio (6) is positive and less than unity for arbi
(z - z 1)/( I -
21 21Z.
with equality holding only if arg z 1 = arg (7)
z,).
z2 =
virtue of inequality (5) as applied to the function f«() = «( - Z1)/(1 the point (= z3, we have '
As one can easily verify, this ratio is invariant under an arbitrary fractional-linear and
I z 1 = 1. This property follows
under fractional-linear transformations, we can always have
z. -Z.
ZI> Z2, Z')=2 --Z ·Z.-24'
1
3 lie on a single circle orthogonal to the circle
Z 1>
Iz I < 1
into itself and that the role of straight
lines in this geometry is played by circles orthogonal to
J." 1
Iz I =
1. Furthermore,
for any fixed z l' the distance D of z 1 and 22 approaches + "" as the point Z2
I
- -------_.
VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS
332
approaches the circle disk
.'=,
Iz I < 1
Iz I
::0
§l.
1. This follows from (9). The set of points z in the
whose distance from a given point Zo in that disk does not exceed
a given quantity p, where p> 1, that is, points for which D(z, zo)
S p, will be
-
INVARIANT FORM OF TIlE SCHWARZ LEMMA
Iz I < 1
Inequality (11) shows that, when the disk
333
is mapped by a function
w = fez) that is regular in the disk and satisfies there the inequality If (z)1 . the hyperbolic distance between the images of two points z 1 and
does not exceed the hyperbolic distance between the points z 1 and
p. It follows from (9) that this hyperbolic disk is defined by the inequality
and is equal to that distance only in the case of a mapping of the disk
.
I
I I-zoz
Zo _.~
+e
11
e!P !p
= tan p
--+
We now present some extensions of the Schwarz lemma.
quality
O. This shows that the l~ear differential element do z in
If(z)1 < 1
in the disk
Suppose that this function has the repre
00
I(z)= ~ cgz k , where n ~ 1. k=n.
I dz I 1-lzI9'
(10)
=
Then, !f(z)1 S \ z In in I z I
I(I
where Idz \ is the Euclidean differential element. The hyperbolic length L of the curve z
I zI < 1.
sentation
the hyperbolic metric is given by the formula
daz =
Iz I < 1
Theorem 4. Let fez) denote a function that is regular and satisfies the ine
14z/ D (Z, z+~Z)=1_I~r9(I+E), 0 as ~z
themselves
piecewise-smooth curve l into a curve of no greater hyperbolic length.
follows from (9) that
->
Z2
into itself. From inequality (12) it follows that such a mapping maps an arbitrary
and, consequently is a Euclidean disk but with center distinct from zoo It also
where (
s ~
in the disk
Z2
referred to as the hyperbolic disk with hyperbolic center z 0 and hyperbolic radius
z-
'-"'.~;;;;;;;;;;;;;;;;;;;5
z (t) for a
StSb
=
< 1 with equality holding only when fez)
::0
(Zn,
where
1.
The proof of this theorem is the same as the proof of the Schwarz lemma if
in the disk
I z I < 1 is defined .as the least upper bound of the sums I ~:~ D (z k' z k +1) for all possible systems of points z k = z (tk), where a = to < t 1 < ... < t n = b. In the·
we consider the function ¢(z) = f(z)/zn, ¢(O) = c n •
case of a piecewise-smooth curve, one can easily show that this length is finite
and satisfies there the inequality If(z) < 1 and has the representation fez) '"
and given by.the formula
Co + 'i'k:nCkZk, where n
Theorem 5. 1) Suppose that. in the disk
I z I < 1.
a function fez) is regular
~ 1. Then, in the disk I z! < 1, we have
Ii
L=~
,
I z' (t) I l
~ ••
dt.
II• (z) I ~
, ..
a
Theorem 3(the invariant form of the Schwarz lemma~ 1) If a function fez) is
Iz \ < 1
and if If(z)1
s1
in that disk, then. for arbitrary points
dar,;;;;, da zm
(14)
with equality holding only when
z 1 and z 2 in it,
l(z)=E (11)
D(j(Zl). l(z~»~D(zlt z~).
a - . 161=1. lal+ 1:
g(z,Pk)
k=l
(2)
everywhere in B. In the particular case in which B is the disk \ z I < 1, this
I'" (f(z» I ~f(O) l-rs, 01-r 1 l+r 1 ()I+f ~lf(z)l~ (O)I_r'
univalently in such a way that
2
I/(z)I~Me k=l
r < 1, we obtain the inequalities
1z I
1 +r
connected domain B bounded by Jordan curves except for poles PI"'" Pv each of which is numbered as many times as its multiplicity. Suppose also that the
< r is equal to 2wor/{1 - r 2 ). In particular, in the
3. Suppose that B is the disk
2
We can also apply the idea of the proof of Lindelof's principle to generalize
=Wo (1 + ')/(1 - ,). Consequently,
all the values assumed ili the disk 1z I < r, where r < 1, by a function f(z) that is regular in B, has positive real part in I z I < 1, and is equal to a positive
the disk l(w - wo)/(w + wo)l
~
< 1 univalently onto the domain G so that
has the form w
341
and having the property that
•
I:ITI ;~:~I'
(3)
k= 1
§4. Hannonic measure. The simplest applications Let B denote a domain in the z·plane bounded by Jordan curves. I) Again 1) The definitions given below and the results remain valid for domains with non Jordan boundaries, specifically, for domains obtained from domains of the type indicated by means of univalent mappings that are continuous up to the boundaries of these domains.
342
VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS
§4. HARMONIC MEASURE. SIMPLEST APPLICA1ltONS
343
let us assume for simplicity that B is finitely connected. Consider a finite num
the first part of the assertion. The second part of the assertion now follows from
ber of arcs (closed or open) of the boundary K of the domain B. Let us denote
the formula w(z, a, B) = 1- w(z, (3, B).
by a the set of all points belonging to these arcs and let us consider a function
The significance of the extension principle consists in the fact that it enables
that is harmonic and bounded in B, that assumes the value 1 on a, and that
us to strengthen various inequalities by replacing complicated parts of the bound
assumes the value 0 at all other points of the boundary K. Here, we exclude the
ary
endpoints of the arcs constituting a since the function is not continuous at such points. By §3 of Chapter VI, this function exists and is unique. Let us denote it by w (z, a, B) and let us call it the harmonic measure of the part a of the bound ary K of the domain B about the point z .. Obviously, 0
S w (z,
where in B. If we partition the boundary K into parts a 1> •• " indicated, we have
a, B)
S
1 every
an of the type
r).k'
B)=l,
Another important property of harmonic measure is given by the following theorem.
Theorem 1.1) Suppose that a function fez) is regular and bounded in a domain B with boundary K and suppose that K is partitioned into two parts a 1 and a 2 • Suppose that all the limiting values of If(z)1 as z approaches interior points of
..
~ w(z,
K with simple ones for which the harmonic measure is easily calculated.
( 1)
arcs constituting the a k (for k = 1,2) through values tn B do not exceed Mk • Then, in B,
k=1
log If (z) I ~ ill (Z,
everywhere in B because the function on the left is a harmonic bounded function in B that assumes the value 1 on the boundary K except at a finite number of
f3 )increases
and when B is extended by notifying only a, the measure w(z, a, B) decreases.
B by modifying {3, for example, we mean replacing the domain B with a domain B' containing B and with bound Here when we speak of extending the domain
ary containing the entire set a. Proof. Suppose that
B' was obtained from B by extending the latter set in
such a way that only {3 is modified. Then, the function II
(z)= ill (z,
cx., 8')
-ill
(z,
(1,
is a
harm~nic function at all points of the domain
which is harmonic and bounded in B, is equal to 0 on a
I
(2)
(3)
B except at the zeros of f(z).
the ak from inside B, all the limiting values of the function u (z)
are nonpo~'4ive. On the other hand, u(z) approaches zeros of f(z). Consequently u (z)
S0
-De
as z approaches the
everywhere in B; that is, inequality (2)
holds and the remark made in the theorem about equality is obvious .. This completes the proof of the theorem. From the theorem just proved, we get
Theorem 2. Suppose that a function f(z) is regular and bounded in a domain B and the boundary K of B is partitioned into two parts at and a 2 • Suppose that the limiting values of
I f(z)1
do not exceed Mk as z approaches interior =
1,2) from inside B. Then, for
every closed set E C B, there exist positive constants '\1 and ,\ 2 independent and is nonnegative on
of the function f(z) such that '\1 + '\2
=
1 and
If(z) I ~M:IM~9
(3. Consequently, this function is nonnegative everywhere in B, which proves on E. 1) Carleman [1921].
(z, ~, B) log MII~
As z apJroaches all points of the boundary K except the endpoints of the arcs constituti~g
points of the arcs constituting the ak (for k
B),
ill
U(z)=loglf(z)\-ill(Z, CXt, B)logMl-'--W(Z,~, B)logMIl
me as ure of its image in B l'
extended by changing only the part {3, the harmonic measure w (z, a,
+
Proof. The function
monic measure of an arbitrary part of the boundary K is mapped into the harmonic
The extension principle. I) If the boundary K of the domain B is partitioned into two parts a and {3 of the type indicated above, then, when the domain B is
B) log M 1
Furthermore, if equality holds in (2) at a point z E: B, it holds for all points in B.
points. Under a univalent mapping of the domain B onto a domain B 1, the har
An important property of harmonic measure is included in the following exten sion principle:
(Xt,
I) F. and R. Nevanlinna [1922].
(4)
~~~~-='-='-=-=-=-------------
344
~--,,:._----------
VIII. MAJORlZATION PRINCIPLES AND APPLICATIONS
---
§4. HARMONIC MEASURE. SIMPLEST APPLICAlIfONS
w~~m=~,
Proof. Suppose that MI and M2 are numbered in such a way that MI < M2. where
1>
is the angle subtended by the segment a at the point z.
2. Suppose that B is the half-disk
If(z) I ~ (~:r(z. «t. B) Mi'
(5)
~
~
For the function fez) in the domain B, we have inequality (2) which, by virtue of (1), we can write in the form
345
I z I < 1,
zs(z)
> 0 and that
a is the
,~ boundary semicircle. Then, one can again verify easily that
2
If we set AI=minze:E~(z, al> B) and A2=I-A 1 , then Al and A2 areobvi ously positive and we obtain from (5)
w(z, a, B) =-(1t -If),
(9)
~
where
If(Z)I~(~:t Mll=M~IM~1
1>
is the angle subtended by the diameter from the point z.
Let us now look at some applications of the theory of harmonic measure to the study of bounded functions.
on E, which completes the proof of the theorem.
Theorem 3. Suppose that a function fez) is regular and bounded in the half
Inequality (2) will find more than one application in what follows. Let us give an example. Suppose that the domain B is the annulus r I
0, and continuous except on the positive half without the point z = 0 of the real axis. Suppose that fez) -> a as z approaches the origin along this half-axis. Then, f(z) -> a uniformly for arbitrary &> 0 as z ---+ 0 in the angle
O:S arg z _log
rfr.
.
_
log
_I I
rfr
1 w(z,abB)-I--/-' ogr r. w(z,%B)-I--/-' ogr r r - z.
1
l
(6)
1
Proof. We may assume that a = 0 and that If(z)1 < 1 in ~(z)
log r/rl ~ Ml~' M lllog'I/'1
interval
log rlr.
(7)
be obtained directly. Specifically, consider the function z(J.f(z), where a is an arbitrary real constant. The maximum value of Iz(J.f(z)1 in the annulus r} 0 in such a way that the point z = 0 is mapped into the point (= 0 and the interval a r is mapped into the positive half of the real axis; We denote this last by a. If z and (are corresponding points, then III
(z, a" B,) = w (C, a, B).
Also, by virtue of (8),
that a is not an integer, then inequality (7) is not sharp. With regard to a sharp bound for M(r), see §4 of Chapter XI. With an eye to subsequent applications
r.
log If (z) I ~ w (z, a" B,) log IS
where r} < r < r2' When we eliminate a, this leads to (7). Obviously,
equality will hold only when fez) = cz -
"
\a
> O. Let us > 0 such that If(z) I :s ( on the
disk Br:l,z\
:s arg z :s
0 in the angle 0
§4. HARMONIC MEASURE. SIMPLEST
this means that fez)
E,
-->
that fez) approaches limiting values a and b as z approaches
0 uniformly as
~. This completes the proof of the theorem.
TT -
along the real axis. Then a = band f(z)
> 0 and l is a Jordan curve lying in that halfplane that terminates at
z = O. Suppose also that fez)
as z
-->
a as z
-->
0 in the angle -~::s arg z ~
Proof. We can again assume that a denote a number in the interval 0
If(z)l~ E
-->
0 along l. Then, fez)
-->
a uniformly
for arbitrary ~> O.
I}
TT -
=
<E < L
I f(z)1 < 1
0 and
in ~ (z)
> O. Let (
Let us choose r sufficiently small that
on the part of the curve l lying in the half-disk BT:
I z I < r,
~ (z)
> O.
Let us denote by a r the part of l extending from z = 0 to the first point at which
l intersects the circle domains
B:
and
I z I = r.
The arc aT partitions the half-disk B r into' two
B;'. Suppose that B:
is adjacent to the negative half of the
B: , we obtain
real axis. By applying inequality (2) to
and let us denote by
a;
-->
00
in the half
f(-1/ z) - b, we conclude that
Proof. If we apply Theorem 3 to the function this function approaches zero uniformly as z
-->
0 in the angle 0 ~ arg z
> O. Furthermore, we can show, in a manner analogous to the proof of Theorem 3, that the function f(- I/z) - a approaches zero as z -~ 0 in the angle ~:s arg z
:s TT.
For
&< TT/2, we see from this that the function fez) approaches
both a and b as z. --> fez)
-->
00
a uniformly as z
in the angle~~ arg z ~ TT -->
00
-
~•
Consequently, a
By means of a conformal mapping, we obtain from Theorem 3 the following two analogues of this theorem.
B:
comple
rays of that angle. Then, a =b and f(z)
-->
a uniformly as z
-->
00
oe
along both
in that angle.
T~eorem 5/1. Suppose that a function fez) is regular and bounded in a simply
real axis. Then, from the extension principle, we have
except
But, as follows from the proof of Theorem 3, the quantity Ill(Z, small r' in (0, r).
o
Thus If(z)!:S ( 1 in the part of
B:
a;:,
:s arg z :s TT - ~
with sufficiently
B;
lying in the sector
r" in (0, r). Suppose now that 0
/I ,
~ ~ acg z
< &< TT/2. If 0
:s
1T
I z I < rl and fi ~ arg z ~ TT -& . This proves the o as z --> 0 in the angle (t ~ arg z :s &. This TT -
0 such that If(z)! that part of
ft
limits ~ and bas' z a!!-proaches c along K from the two sides. Then, a
B;)'~(J)(z, a;" Br)~(J)(z, a;, B T )·
0
band
completes the proof of the theorem.
angle. Suppose that it has de finite limits a and b as z approaches
the portion of the boundary of the domain
I z I < r',
=
through values in the halfplane ~ (z) ~ O. This
connec~ed domain B bounded by a Jordan curve, that it is continuous on
positive lower bound olin the sector
~,
TT -
where ~
B r complementary to fJr' Finally, let us denote by 0;' the interval (0, r) of the Ol(Z, aT'
:s
Theorem 5'. Suppose that a function fez) is regular and bounded in the angle
Let us denote by fJr the portion of the boundary of the domain ,
a uniformly as z
and +00
~1 < arg z B:). mentary to a r
-->
-00
347
plane ~ (z) ~ O.
Theorem 4. Suppose that a function fez) is regular and bounded in the half plane g(z)
APPLI~ATlONS
1) Lindelof [1915]. 2) Millou" [1924].
. l-Iz"
points on
1
( 1 1)
~===~~:o,,~
348
VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS
For any Zo in the disk
Iz I < 1,
there exists a domain B containing Zo and in it
'0
mapped into a point ~o,
I I < 1.
half-disk B
axis. Specifically, let us draw through
into itself, we can arrange to ha ve the point
'0
'0
lie on the imaginary
a circle orthogonal to the circle
1"
= 1
and to the real axis. Denote by a the point of intersection of this circle with the
B, we
real axis. When we map the upper half of the '~plane into itself in such a way
obtain
'log If (z) 1 ~ Ol (z, ~, B) log e.
(12)
Here, the question consists in finding a bound for the harmonic measure w(z, {3, B). We shall first need to prove an inequality for univalent functions. Suppose that a function g(O is regular and univalent in the disk I~I
1- viZl ;;0--
O. We denote by z = g (,) the inverse function providing the mapping mentioned. We
I
710/0 - 710]2
in (13) and integrate with respect to r from 0
to r, we obtain, for arbitrary ~ in the disk
above is mapped into the imaginary axis. In particular, the point
Ol
4
Equality holds in (13) in the case of the function g(~) where \ 711
tively, the half-disk B ' is mapped into itself and the orthogonal circle mentioned
extended beyond the real axis, and it maps -the complete disk
g(C+Co) 1+(oC -g(C o) h (Q= .
that the points -1, a, and + 1 are mapped into the points -1, 0, and + 1 respec
have Zo = g(ito)· The function g(~) can, according to the symmetry principle, be
not assume the value zero in that disk. Then, the function
!g(O)!
349
By means of a supplementary mapping of the
z = zoo
I
~~~~~-_.==--:-=~~~=-
SI~PLEST APPLICATIONS
§4. HARMONIC MEASURE.
the function fez) is such that equality holds in the first of inequalities (11) at
Proof. I) When we apply inequality (2) to the function fez) for z €
===
---,~=.
f3,
B). We map the domain B
B ': I~\ < 1, 15 (~) > 0 in such a way that the arc a is mapped into the boundary diameter of the half-disk. Suppose that the point z 0 € B is
onto the half-disk
lead to the first of inequalities (11). Equality holds in it only when
B is a disk
with cut along a suitable radius and the function fez) is such that, as z assumes values on the half-disk B I, the function log' f(z) I is a harmonic function that is equal to 2 (1 - ¢/17) log ( on the boundary, where ¢ is the same as in formula (9),
1) The proof given here with the sharp inequality was given by E. Schmidt [1932].
From the inequality
§5. ON THE NUMBER OF ASYMPTOTIC VAL"ttES
VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS
350
351
Consequently, for fixed z and sufficiently large r,
arc
. I-izi 1-lzl 1-lzl I + I z I?: 1 + I z I ~ 2
Slll
(J)
we get the second of inequalities (11). This completes the proof of the theorem. Now, we shall finally give an application of the theory of harmonic measure to a single question assoch ted with the principle of the maximum modulus. We
(z,
(x,
B) r--.J ~ (-y-
, Zo
=
which shows that, for suitable A
in
~ (z) > 0
I,;
or (2) there exist pos itive numbers A and R such that
for r>
I < r, 15 (z) > 0
and let us denote
by a its boundary semicircle. From inequality (2), we have
log If (z) I ~ (J) (z,
(x,
B) log M (r).
(18)
B) = ; -
['It - (; (; -
arc tan r Y
arc tan r
r
X'
M (r)
I in the entire angle or (2)
>e
Arl a /
(20)
where! M (r) is the maximum value of If(z)1 on an arc of the circle
I
z
1 =
r lying
R.
I
§5. On the number of asymptotic values of entire functions of finite order
Let us give an application of the theory of harmonic measure to a more com plex question, one dealing with entire functions, Suppose that fez) is an entire function and that l is a continuous curve in the z-plane that extends from some finite point to 00,
00.
If, as we let z move along
the function f(z) approaches a definite limit a, then a is called
the asymptotic value of the entire function fez) defined by the curve l. Suppose
1.
x )] = : (arc tan r Y x
:S
there exist positive numbers A and R such that
l towards
But by (9), if z = x + iy, (x,
r
with vertex at z = O. Suppose that the limiting values of 1f(z)1 do not exceed
\
R.
Proof. Let us denote by B the half-disk \ z
(z,
log M (r)
1m
> 0 and sufficiently large r, inequality (17)
in th1 angle in guestion, where r>
(17)
where M(r) denotes the maximum value of If(z)\ on the semicircle Iz\ = r,
(J)
TTU
and that all the limiting values of its modulus are bounded above by I
M(r»e Ar
15 (z) > 0
l'
r ..... 00
1 at finite boundary points. Then either (1) If(z)I
as z approaches any finite point on the real axis. Then, either (I) 1f(z)\ :::: I in
(19)
Theorem 7'. Suppose that a function fez) is regular in an angle of magnitude
> O. Now, we have
15 (z) > 0
4y
1tr'
theorem that will be used in the following section.
> I in the halfplane ~(z) > O. The reason for this is the fact that the limit
Theorem 7. 1) Suppose that a function fez) is regular in the halfplane
r--.J
By performing a conformal mapping, we obtain from Theorem 7 a generalizing
is regular in the upper half
plane and its modulus is equal to I at finite points of the real axis. Furthermore,
~ (z)
2
holds. This completes the proof of the theorem.
violated even at a single point of the boundary. Then, the modulus principle no
= oe
If (zo) I ~ 4Yo
different when the requirement of boundedness of the limiting values of If(z)1 is
ing values of the function are not all finite as z approaches the point z
2
Xo + iyo denote a point in B such that If(zo)1 > 1. Then, from (18) and (19), log
the domain B, then If(z)1 :::: M throughout the domain B. The situation is quite
If(z)1
+ _Y_) = ~1t ~ r+x r _x
we have
values of its modulus exceed M as z approaches any of the boundary points of
iz
r-x
Suppose now that the inequality If(z)1 :::: I does not hold everywhere in B. Let
recall that, if a function fez) is regular in a domain B and if none of the limiting
longer holds. For example, the function fez) = e-
1t
+ arc tan r +- x ) ,
that two curves II and lz of the type described define finite asymptotic values al and az. If al
t- az,
these values are considered distinct; if al = az, then let
us agree to consider the curves II and lz as defining distinct asymptotic values only in the case when there is some R > 0 such that the curves II and lz can not be connected by a continuous curve lying in th~ domain - - 1 ) Phragmen and Linde lof [19081.
Iz \ > R
and having
-------_.. _--"'-_. __ ._ ..._..'-
=..;;;,;;;~
__ __._- ._--_.
VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS
352
-
,,--------
.._ _._.
--~--------~~~
__ _ _-_ -----------_._..
.. _._._ ...- .... _... _--_._-"'.,,--_ ....-.__ ...
... _-- _.'-'-
§5. ON THE NUMBER OF ASYMPTOTIC VALUES
the property that the oscillation of the function fez) on it is arbitrarily small. In
finite asymptotic values.
the theory of entire functions, this convention identifies the asymptotic values of the entire function fez) with nonalgebraic (transcendental) singularities of the inverse function.
If an entire function fez) is such that, for some finite k > 0 and sufficiently
353
Proof. We note that if two curves II and 12 that extend to infinity define two distinct asymptotic values of the function fez), then, outside some disk
Iz I s
R they have no points in common. Therefore, we can in general assume
them to have no points in common. Without changing the asymptotic values, we can
large r, we have M(r) ""
If(z)I>N=
sup
If(z)1
11' •••• In
is nonempty. This sets consists of domains that extend to infinity. This is true because, if one of them were bounded, inside it we would have If(z)1
> N, which
the maximum principle asserts is impossible. For every k, we shall consider not
B k but one of the domains lying in Bk just mentioned and we shall now denote I z I = ro contains
it by B k. Let ro denote a positive number such that the circle I)
AhHors [19301
2) Goluzin [1946dl
points belonging to all the domains B k, for k = 1,' .. , n. The maximum value of
354
VIII. MAJORIZATION PRINOPLES AND APPLICATIONS
If(z)! in the part of B k contained in the circle \ z
I=
'0
I z \ S'o
§5. ON THE NUMBER OF ASYMPTOTIC VALUES
is attained at some point Zk on
Let us find an upper bound for the harmonic measure w «( k>
which does not lie on the boundary of B k' Let us now look at
I z I > '0'
the part of the domain B k lying in the domain
It consists of domains.
We denote by B k that domain containing the point z k on its boundary. This B k must extend
to
00
because otherwise the maximum principle would again lead us
to a contradiction. If uk-is the longest arc of the circle
1
z I = '0 that contains a
point z k and constitutes part of the boundary of the domain B ing the domain B k with the domain B~ consisting of B
k'
k'
then, by replac·
uk> and that domain
respect to the sector obtained from lk by adjoining suitable points of the disk
,,= (
1('
< '0' It follows from the mapping of the sector by the function l/&k , where &k = (¢k+l - ¢k)ITT, that this last measure is equal in the half-disk 1(' l.s r* = " li/&k,g «(') > 0 to the harmonic measure of the boundary semicircle I" 1= r*, 0 < arg (' < 77 relative to some point = ~k + iTJk on ,(' I = ,J/&k; that is,
a
in accordance with (9) of §4, it is equal to
about the circle I z I ='0 which is adjacent to uk' Since z k is an interior point of the arc uk> then, if we denote by N k the maximum value of If(z)1 on the boWld· for all k
=
BL
we have N k
< I fez k)l.
I '1jk -2 (arc tan 1t r* - Ek
(0
(C k ,
rJ.k'
Ok)
~ 3.. (arc tan 1t
'0
l/(q 1), so that (p + l)/p < q, we have (p + I)A nk < PA nk +l' In this case, the serie~ (4) is a grouped series with respect to the power series representation of
a power series of the form (1) with radius of convergence given by the equation
R (1 + R) = 1, so that R = (y5- 1)/2. For this power series, the series (2) is a
(p+1»)'
.k
Here, the two quantities constituting the subscript to
n=O
I z (1 + z)1 < 1.
, k - l , 2, ''',
z
this property is called the hyperconvergence of the power
grouped series and it converges on the set
nh+l_
1 at which fez) is regular. 1)
changing their order, this series converges on some set of points z outside the
where the An are integers such that AO
(3)
Proof. Suppose that the function fez) is regular at the point z = 1 (if it is regular at a point z = e i '\ we could consider the function f(e-iaz». Let us set
Suppose that we have a power series
s R,
(z),
k+1
converges in a sufficiently small neighborhood of an arbitrary point on the circle
Iz I =
theory of power series. Let us now look at the question of what is known as
1z I
An
k
PAnt (z) = CoZAo
2p, which are defined by the same rays.
§6. The hyperconvergence of power series
closed disk
357
00
the rays arg z = kTT/p, for k = 1, 2, •. " 2p; the second has 2p asymptotic values =
_
1, 2," " where q> 1 and is independent of k, then the grouped series
of order p. The first of these has 2p distinct asymptotic values 0 defined by
e21Tkil P, for k
"__ ··_·
§6. THE HYFERCONVERGENCE OF POWER SERIES
z
sin zP zP an
"
.•.
~:;:;;;,:,'''''''. qA nh, k
1) Ostrowski [1922].
.
1) In particular, if the inequality An +l > qA n , where q> 1, is satisfied for all n = 1, 2,"', then the power series itself can be regarded as grouped. In this case, PAJJ'l.n+l(Z)=
c n / n + 1. In accordance with Theorem 1, it will therefore converge in a sufficiently small neighborhood of every re gular point on z = 1. But this last is impossible because a power series cannot converge outside its circle of convergence. This means that, in the present case, the function f(z) cannot be continued beyond the circle of convergence. This is a familiar theorem of Hadamard.
I I
... ."~
... "d.!,_
358
"' o. But u*(zo) - v(zo) = u(zo) - v(zo) and, in a sufficiently small neigh borhood of zo, we have u*~z) - v(z) ~ u(z) - v(z). This means that the difference U(z) = u*(z) - v(z) has at the point Zo a positive maximum with respect to some neighborhood Iz - z 0 I < ro of the point z 0 and in that neighborhood it is positive. Let us show that this last is impossible. The relations ~u* ~ 4e 2u * ,~v = 4e 2v , and u* > v imply that, in the disk Iz - zol < ro,
A. (u* - v):;:: 4 (e'iu* - e'i'V)
an arbitrary function w
that is regular in \ z I < 1. How
= f(z) = z + C2z2 + •••
ever, it follows from Theorem 7 of §7 of Chapter II that there exists a positive
Iz I < 1
under an arbitrary function of
covers, for example, some half-disk with center at the origin and radius independ ent of the form of f(z). Bloch has proved more than this: there exists a positive number p such that, for an arbitrary function w = f(z) = z + ... that is regular in
, I z I < 1,
the w-plane contains a disk of radius p that is a one -sheeted image of
some portion of the disk
Iz \ < 1
under the mapping w
=
f(z).
Let us now introduce some new notation. We denote by L f the radius of the largest disk in the w-plane that is completely covered by the image of the disk
> 0,
\ z I < 1 under a function w
= f(z) = z + ••• that
is regular in
I z I < 1.
We denote
by B f the radius of the largest disk in the w-plane that is the one-sheeted image that is, ~U > O. On the other hand, in accordance with Green's formula we have for r
< ro ,
Iz-t=r ~~ rdB= lzJJO,
of a subdomain of the disk , z
=
which the function w
=
f(z) maps the disk
B= inf Bf ,
Zo + re;/3; that is,
f (z)
2x
iJ
\
.
7fi ~ U(zo+re'~dB>O. If we now integrate with respect to r from 0 to r, where 0 2"
U(zo)
~ 2~ ~
U(zo
same mapping. (In other words, B f
is the radius of the largest one-sheeted disk lying on the Riemann surface onto define
where we set z
I < 1 .under the
Iz I < 1
bijectively.) Furthermore, we
L= inf Lf • f
(z)
Here'lhese infima are over all functions f(z) = z + '. •. that are regular in the disk I z I < ; here B is called the Bloch constant l ) and L is the Landau constant.
< r < ro,
we obtain
The n merical values of these two ·consrants are as yet unknown: we have only upper and lower bounds for them. 2 )
+ rele) dB.
Tbe theorem proved above enables us to obtain the best results as regards lower bounds for the constants Band L. We begin with a lower bound for B.
Since the function U(z) attains its maximum at the point zo, this last ine quality is possible only when V (z) == U(zo) on Iz - zol = r for all r in the inter val 0
< r < ro; this would mean U(z)
== U(zo) = const in Iz - zo!
< roo This is
incompatible with the inequality ~U > O.
R -> 1, we see that A(z) :::; 1/(1 - \ z
= f(z) = z + .•.
is regular in
Iz I < 1
and has
r'
finite B f' Let z 0 denote an arbitrary point in the disk I z 1 < 1. If (z 0) t. 0, then the inverse function z = f- 1(w) is regular in a neighborhood of Wo = f(zo)· We denote by p(wo) the radius of the largest circle with center at Wo in which
Thus, we have shown that u(z):::; v(z) 2 1 ),
Suppose that the function w
=
log [R/(R 2-/ z \ 2)] in
I z 1< R.
As
that is, that ds :::; du. This completes
the proof of the theorem.
the function z =
f-
1
f- 1(w)
(w) is contained in
is regular and the image of which under the mapping z
I z I < 1.
=
Obviously, B f is an upper bound for the numbers
p(wo) for all possible Zo such that \zol
< 1.
Now, following Ahlfors, let us give an application of this theorem to covering questions.
In §7 of Chapter II, it was pointed out that there does not exist a disk Izi 0, that is completely covered by the image of the disk \ z 1 < 1 under
I) Bloch [1925] showed that B> O.
2) On the ba sis of elementary considerations. Landau [1929] gave some numerical
bounds for Band L.
_
--~~~~-==="---~----:::C-C:_'.=-"'"
364
---_.- -- ----
VllI. MAJORIZATION PRINCIPLES AND APPLICATIONS
If we denote by Gzo the domain in I z I < I that is mapped by w = [(z) onto < p(wo), then Gzo must contain on its boundary either points
I or points at which f' (z) = O. Furthermore, we can easily see that p (w), where w = [(z), is a continuous function of z in I z I < I and we
Iz I =
have, close to the point zo, where Izol where Wo = [(zo).
< I and
Let us now define a metric ds = A(z)1 dz
A(Z)=
I
in
f' (zo) = 0, I z I < I,
p(w) = I w - wol,
taking
function yt(A - t) increases in the interval 0
V'A 1 nCn (~ - zo)IJ-1 + ... I 2I cn(z-zo)n+ ···1 Ill(A-lcn(z-zo)lJ+ ... Zo
1%-1 2Jcn
+ ... I
'VAlncn
II
+ ···1
»(A -I cn (z - zo)n
0< t
ds
~
=
+ ... )
/
[Ca»! A,
yiel~s Bf
=
.
~ --.!!J. II ,(z) I ~~t everywhere in I z I < 1. In
fez). Furthermore, throughout the entire
particular, with z = 0, it
2: v'3/4. Since the function fez) is any function that is regular in
va
B~4=0.43
....
In an analogous manner, we can find a lower bound for L. For every z 0 such
Iz 01 < 1,
let us denote by p(wo) the radius of the largest circle with center
= f(zo) that is completely covered by the image B of the disk I z I 3Bf , we have
I
•
that
cal
< t < Bf , if we
zr'
VAlf'(z)I~2~(~-:Bf)
.
< I and ['(zo).j,. O. On the
w=!(z).
1
i)
VA If' (z) I
p* (w)=! w- lea)!,
< A/3. Therefore, if we take A > 3Bf' we have in I z I < 1 the inequality
obtain
f'
2
0, as was shown is regular and
do, that is,
Since 0(A - t) increases in 0
boundary of Gzo is a point a which either lies on the circle I z [ = 1 or has the properties that (a) = 0 and I a I < 1. In Gzo ' we define the metric ds" = A*(z)ldzl, where ( )
I
tions of the theorem established above are satisfied. But Vi(A - t) increases for
,I
). '" Z
r' (z) =
What has been said leads to the conclusion that if A is such that the function
=V'A=A:-,-,If,-'(,,-,z):..J.,.I_
n> 2, we have A(zo) = O. (Here, A(zo) is defined as
Now, let us consider the point Zo such that Izol
Consequently, when this
or else the metric ds = '\(z)! dz
2 Y'P(W) (A - p (w» ~ -1--1
This shows that, for n = 2, the metric in question is continuous and, in fact, regular at Zo and that, for lim z~zo ~ A(z).)
=0
n>l,
in a neighborhood of it, then, in a neighborhood of zo, we have pew) I[(z) - [(zo)1 and, consequently,
_I Z -
< t < Bf'
v't(A - t) increases in the interval 0 < t < Bf' then, for the metric (4), the condi
00
-
365
last condition is satisfied, the metric (5) is a support metric for the metric (4) and
(4)
r'
A(z) :.-
~~~~'=~
hence can, itself, serve as support metric.
at all points of the disk I z I < I at which (z) .j,. O. On the other hand, if a point z 0 such that Iz 0 I < 1 has the property that
cn:;cO,
=-
it satisfies condition (I). With regard to points at which
where A is a constant exceeding Sf. The function A(z) is obviously continuous
+ k=n ~ ck(z-zol,
-';"--
domain Gzo ' we obviously have p*(w) 2: p(w) and p* (wo) = p(wo). Therefore, in Gzo we have A* (w) ~ '\(w) and ,\* (wo) = '\(wo) though only on condition that the
above, at such points ,\(z)
V'AIf' (z) 1 ,W=!(z), 2 'JIP'(W) (A- p (w»
!(z)=!(Zo)
~-_.
§7. NONANALYTIC GENERALIZATION OF SCHWA~Z LEMMA
the disk Iw - wol of the circle
------
=
fez) and let us define on
Iz I < 1
the metric ds = A(z)ldzl,
where
}.(z)=
1/'(%)/
c .
w=!(z),
2p (w) log P (w)
C being a constant greater than
L f • I ) This metric is continuous in I z I < 1.
1) Again, it will be sufficient to consider functions fez) with finite Lf.
(7)
i
366
VIII. MAJORIZATION PRINCIPLES AND APPLICATIONS
Now, let us denote by Gzo the largest domain contained in
1
z
I
1, we see that \ A ~ (1)1 ~ n. Now from
f(."k z )
e 21Til for n = 1, 2,"', it follows that B. But this means that the function where."
k=O
II! S+1 rIdC' I n-I
Proof. Let us consider first the second case, in which the image B of the under the mapping w
21ti, 1,1=1
. [~ Ck ]! An (0 "'"""f'JiiT dr:
n-I
ordinate to F(z). Suppose that the image of the disk I z I < 1 under F(z) is a starlike domain. Then, we have the sharp inequalities I an \ ~ n, for n = 1,2, ....
I zI < 1
n-I
r ,J
1
and, consequently,
lar in I z I < 1. Let F(z) denote a function that is regular and univalent in the same disk. Suppose that F (0) = 0 and F '(0) = 1 and suppose that f(z) is sub
disk
373
r
the first of inequalities (12) only for the function F(¢) = ¢/(I - "'1¢)2, where
1"'11
'''''''''-''.,~
§s. MAJORIZATION OF SUBORDINATE ANALYTIC fuNCTIONS
VIII. MAJORIZATION PRINOPLES AND APPLICA nONS
Since I¢(z)! ~ I z
~.,-"_.~_ •.;~~.~~,.,,-~
disk I z 1~ \ Zo I, containing the disk I z I ~ I Zo 1 , and bounded by an arc of the circle I z 1 = I zol 2 and the arcs of two curves passing through Zo and tangent to 2
the circle 1 z I
= I
zol 2. Then Wo
=
¢(zo) belongs to ~zo·
Proof. In accordance with inequality (3) of
f(z)
=
z-I¢(z), the point Wo
=
§1
as applied to the function
¢(zo) lies in the disk with boundary defined by
1) This inequality can also be proved if we replace the assumption that the image of the disk I 1 < 1 under the mapping w = F(z) is starlike with the assumption that it is symmetric about the real axis. For more infomlation on this, see Littlewood [1944].
z
~
=~~~U~"'~,?"-,-":~~:~~~~E:'~'0_",,",~"~"-~""~~~-~~;:·:!:!"':~~F=~~~I:-y'J'~::-~~.'~~_~~'::~
~
374
VIII. MAJ ORIZA nON PRINCIPLES AND APPLlCA nONS
the equation I(w - a 1z 0 )/(zo - a 1w)!
=
§s. MAJORIZATION OF SUBORDINATE ANALYTIC "UNCTIONS
I zol, where a 1 = ¢ '(0) and 0 sal S 1. as 1, let us find the envelppe of all
If we now take a= a 1 in the interval 0 S
the disks obtained. These disks are obviously contained in the disk Iw and their centers lie on the segment (0,
1
s
I
Zo
I
at all points z ~ z 0 in the domain ~z 0 mentioned in the lemma. This is true because the function F' (f(z))
=
a 1z + .•• satisfies the conditions of the lemma
and hence, by ,applying (16) to the point z = F -1([(zO)) € ~zo, we obtain (15).
The equation of the envelope is
Z1»)'
To prove inequality (16) with 1z 0 I
determined from the equations
=
X,
we define the number p € (0, 1) as
a root of the equation
F (w, IX) = ffi -
(lOg w- azo ) Zo-aw
F~(W, 1X)=ffi(, w-az Zo
w
_
o
= log 1Zo I, zo-aw
(l
)=0.
(14)
o
aw)
) =0
m(wzo-aw - azo ) =
or
IF (z) I ~ I F' (0) I II ~~ ~ 1\1
I z I < 1, ,
we have
IF (zo) I> IF' (0) 111 !z~ ~
S
as 1,
this determines the arcs of the
Then, on the boundary of the domain ~zo there exists a point z 1
at
Z,
spo nding to the value s a = 1, 0, and - 1) and the other pas sing through the points
arc by 1. Theiangent to the image of t he arc l under the map
S a 1 S 1,
=
cover the domain ~zo mentioned in the lemma. This completes
the proof of the lemma. Theorem 6. 1) Let fez) and F (z) denote two functions de fined and regular 1z 1
in the disk
Izol'i= 116 ,
for z € ~zo.
conjugate, the equation takes the form
m( (w - az_)1(zo -
p p)1
Since the function F(z) is univalent in
If we replace the first fraction in the second of these equations with its complex
Izo
375
it will be sufficient to show that
I F(z) I 3 -..,(8, this derivative is negative for a= 1.
From this we conclude that the inequality
+1 rr
2
I 'Pt I~) = (r + I
(1 -
,! !)
..,(8 = (l
-=- r I 'ill/).
Theorem 7 be univalent in in the disk
+
a+r
,!
Iz I < 3 - ..,(8
l+ ar
'()I a(l+r l )+2r 1-1,11 1'P Z ~ 1 +rl+2ar l-r 2 ' If we multiply (21) and (22) and set a
I I' (z) I F' (z)
[_
~
(22)
= (l
+ r 2 )/2r, a ~ 1, we obtain
(
0 -
{1.)2
0-
1
+I +
011
0
11
(23) •
Here, the sign- of the derivative with respect to a of the right-hand member coin cides with the sign of the polynomial
p (a) =a3 - 3a - (3a~
+ 1) a. -
2aa.~.
1,
p(a.) ~p(I)=a3 - 3a~ - 5a -1 =(a+ 1) (a~ - 4a -1), -
4a - 1 ~ 0, that is, if a ~ 2 + ,/'5, we have p ( a) ~ 0 in 0 S
a
I z I < 1,
~ I 'Pt I
f' (r)
(a+a)8'
379
(z tt:z), 0 ~ a. ~ 1,
which satisfy the conditions of Theorem 7. We have I) Goluzin [1951rl.
I z I < 1,
then the inequality
If' (z) I s IF' (z) I holds
though not always in a larger disk. 1 )
§l.
LIMITING VALUES OF POISSON'S INTEGttAL
381
Theorem 1. If the function fee) is continuous at e = eo' then u(r, e) - .... ieo by any mode of approach through
f(e ) as (= reie approaches the point e o points in the open disk 1(/ < L
(>
Proof. For given CHAPTER IX
If(t) - f(e o)! since
BOUNDARY VALUE PROBLEMS
FOR ANALYTIC FUNCTIONS DEFINED ON A DISK
< d2
0, we take 0 in the interval 0
whenever
It - eol < O.
< 0 < 17
Suppose that Ie -
such that
eol < 0/2.
Then,
2"
;~ ~
1)
P(r, t-B)dt=l,
(3)
o we have
§1. Limiting values of Poisson's integral
U
As we know, an integral of the form
;1t
(r, B) -- f (B o) = .
2"
~ (f (t) -
f (B o» P (r, t - B) dt.
b
Let uS partition the integral on the right into two integrals II and 1 , the fiest 2
·2"
u(r,
B)=2~ ~f(t)p(r,
over ~he arc (eo - 0, eo + 0) and the second over the remaining arc 11 of the
(1)
t-B)dt,
.circle\\(1 = 1. Noting that Per, e)
where
\
! rcos 0+ r1I=1+2
00
l-r
?
(2)
n
r cosnB
Since It -
e\ > 0/2
is called Poisson's integral. t and has period 217, the integral (1) yields the solution of the classical Dirichlet
in the open disk \(1
< 1,
where (= reie. This solution is a harmonic function
it is continuous in the closed disk \(\ :5 1, and it as·
sumes the values f(e) on the circle
1(1
= L In this section, we shall study Pois
son's integral under less stringent conditions imposed on the function f(t). Let f(t) denote an arbitrary real function that is summable in (0, 217) and tbat has period 217. Obviously, the function u (r, e) defined is again a harmonic function in the disk
1(\
< 1.
2)
by the integral (1)
Then we have
on 11 and hence, P (r, t -
Il~l~
In the case in which a real function f(t} is continuous for all real values of
I( I < 1,
j
2"
E~ P(r, t-B)dt=z. E P(r, t-B)dt~~
°0- 0
,
11=1
problem for the disk
°0+0
E\
l
P(r, B)=l -
\1 ~hl~4~
> 0 in the disk 1(\ < 1, we obtain
l - rll
~
1--2rcos-+rll 2
This shows that 12
-->
0 as r
--->
e) < P (r, 1 \
·21tjlf(t)!dt. 11
L Consequently, there exists an TJ> 0 such
1121 1 - TJ· Consequently, for r> 1 - TJ and Ie - e01< 0/2, we have 1u (r, e) - fee 0)1 < (, which completes the proof of the theorem.
that
As a consequence of Theorem 1, we obtain the familiar theorem that, if the function f(t) is continuous foe all t, then the function u(r, e) defined on the circle
/(1
=
1 as the limit of u(r, e) when the approach is through points inside
the disk I( I < 1 is continuous on '(I :5 1 and u (1, e)
if fee) is continuous on some arc of the circle 1(1 1) Stricdy speaking, this chapter is not a parr of the geometric theory of functions. However, the information contained in it will be used to a considerable extent in the fol lowing chapter. 2)ln this and the following chapters, a familiarity with the fundamentals of the theory of functions of a real variable is assumed. Thus the integral (1) is understood in the sense of Lebesgue. Also, because of the periodicity of !(t), we can displace the interval (0, 217) and hence we can interpret the integral as being over the circle 1~ I = 1.
380
0/2), we have
=
=
f(e). On the other hand,
1, then the function u(r, e)
defined on that arc as the limit of u (r, e) when the approach is through points inside the disk 1(1 < 1 will be continuous in the union of the open disk set of interior points of that arc.
1(1 < 1 and the
Theorem 2. If a function fee) has a finite derivative ~t e = eo' then au(r, e)/ae -+ ['(eo) as (= reie approaches the point e,e o through points in
-:--=::===-~=~~~=-~:':~,,=,~~::~i'>;.~~,;';':~~;~';'~i;='~;:":;'-~;0=~:Mi:iw~~..:i~;:~,:;:~_~))~~~"";;_,,,;i.~~'i'';;;;"_';;;~~''i;;;,.-.i'~
382
IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS
the open disk
1(1
side the disk
I(I < O.
§1. LIMITING VALUES OF POISSON'S INTEGR'AL
< 1 along any nontangential path (that is, a path that lies in h ' l y in . h some ang 1e w£t vertex at e iB o tat, sUI,(f"£c£ent 1y c 1ose to e 'iB o, l'£es enUre
linear in the interval (a, a + 217): [(()
that is
A() + B and that is extended as a periodic
=
2r (t-ll o) sin (t-6) I elf --: reID j II
Q But le it - reiBI ~ !ei(t-B)_
Proof. Let us fitst prove the theorem for the case of a function [(()
rl > Isin{t- ()j
,ou (r, 6) -1 all
2~
-
~
a6
2~
-
a
1
=-~(f(a+27t- 0) -
2r I t-O o I
I Q I~ I eU -
C I(t) oP (r, t -,ll) dt
I
~
at
l(a+O» P(r, a -a)
'0 i (Oo+~+~) re''0 =elo+se 2" we have •
It follows that, for a < () 0 < a + 217, the derivative au (r, ()/ a() --> A as re iB iBo iBo along any path leading to the point e from inside < 1. e
' . I re,91 = \ ell - ei90 - e;e
--->
1(1
= \ 2e
~~~_~
k
6)+r 9
i (
) /-9 0 --CL
2
t-ll
sin - 2_0 -
&
---> [
I
'B as reI
(() 0)
e
--->
l
membering the periodicity of [(()),
all
6) _
aUI (r, oil
ll)
e; I~ 2] sin
(ex -
t-ll --T) I
a0 I
..
= _1
sm'2
Q is bounded for It - ()ol < a o' If re iB is sufficiently close to e iBo , that is, if ( < (0' then Q is bounded on the remaining parr of the interval a < t < a + 217 because the denominator in (6) is then bounded below. Thus, for « and all t, we have 1QI
a+2" \'
2~
~ a
(I (t)t -- I6 (llo) _ 0
XP(r t-a) ,
I' (a »)
au (r,
0
I
2r(t-llo)sin(t--ll) 1- 2r cos (t - 6) r9
+
dt,
We write this factor as follows:
< a 0 < 17/2,
00
ll)
_,aul
< Q0'
(r, 6) all
with the tan
(0
Therefore, we obtain from (5)
I.,;:: Qo -= 2~
(5)
Let us show that the las t factor in the integrand in this equation is bounded iBb for t € (a, a + 217) if re iB is sufficiently close to e and lies in an angle with vertex at e iB o and sides making angles, a 0' where 0 gent to the circle 1(1 = 1 at the point e,B o.
~
that is,
iB o
Now, from (4) and the same relationship applied to u (r, ()) we obtain, re
au (r,
I
QI~~,
f' (()
())/ a()
a;o~a;-S;7t-I1o,
(again the same sort of substitution), COllsequently,
e ,B o and extended as a periodic function outside that interval. From what we
au 1 (r,
- () 0 1 ::; a o and note that
(4)
•
Let us denote by u (r, ()) the function obtained from (1) if we replace [(t) with l the linear function [(() 0) + (t - () 0) 0) in the interval (a, a + 217), where e ai l have shown,
2
1i
' -t -26-0 I~ O,
. (0o+~+-")
,
l ell -
a
ou(r, 6)=_1 \/(t)p(r, t-a) 1-2rcos(t
re lO I .
Furthermore, if we assume for the moment that
a
a+2" 1 C +~ ~ 1'(t)P(r, t-a)da=-AP(r, a,-a)+A.
all
(this last is obtained by replacing
aginary parr). Consequently,
a+2"
C I(t) ap (r, t - 6) dt
(6)
the absolute value of the expression as a whole with the absolute value of the im
function outside that interv_al. Then, from (1) we obtain a+2~
383
a+2" C I I(t) - 1(°0) -I' (a) IP(r t - a) dt ~ t -llo 0, • a
But Theorem 1 is applicable to this integral because
f' (() 0)
exists and hence the
first factor in the integrand approaches 0 as t,---> () o' Thus, this integral itself approaches 0 as (---> 0, that is, as re iB --> e' Bo in the angle in question. Therefore, by passing to the limit, we have , I 1m
ou a6(r,
ll) -11'm aUI (r, ll) =1' (a) all 0,
384
IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS
which completes the proof of the theorem.
repeating the same reasoning as above, we can prove
Theorem 3. If the value of the function f(()) in (1) at (j equal to (j=(jo:(d/d())rtf(t}dt=f«(j)18
0
= (jo
is finite and
Theorem 4. If the function a (t) in (7) has a finite derivative at () = () , -8 0 where < () 0"< 2TT, then u (r, ()) - a' «(j 0) as (, = re i8 approaches e' 0 along
°
then u(r, (j)-f«(jo) as (,=re i8
approaches the point e .8 0 along an arbitrary non tangential path from inside
any nontangential path in
1(,1 < 1.
1(,1 < 1.
Since aCt) has a derivative almost everywhere in (0, 2TT), we obtain, as be
Proof. Suppose that 1) 0 interval a
385
§2. REPRESENTATION OF HARMONIC FUNCTIONS
< (j < a + 2TT
-
< a < (j o'
2TT
We define the function F «(j) in the
fore, Corollary 2. The Poisson-Stieltjes integral (7) has almost everywhere on
by the formula
I(, I = 1 limiting values equal to a' «()) along all nontangential paths.
9
F (lJ) = ~ f (a} de,
All the theorems of this section are due to Fatou,
a
I
F' «() 0)
=
f(() 0)' If we now inte §2. T~e representation of harmonic functions
Ia+2,.
by means of Poisson's integral and the Poisson-Stieltjes integral
u(r, e)=2~F(t)P(r, t-e) a
a+2" __ 1 2~
\
~
n
at
~
2~'
tegra. The answers to these questions involve special classes of harmon.ic func tions ~which we shall introduce special notation. Specifically, we denote by
•
a+~
this section, we shall take up questions of the possibility of representing
harm nic functions by means of Poisson's integral and the Poisson-Stieltjes in
F(t) ap(r, t-O) dt=~P(r a-e)
a
who initiated the in
vestigation of the boundary properties of analytic functions.
and then extend it as a periodic function. Then, grate (1) by parts, we obtain
1)
X ~fOO~+~~~FOOP~t-~~
h , where p > 0, the class of functions u (r, (j) that are harmonic in the disk p
\(,1 < 1 and have the property that the integral 2"
~ III (r, e) IP de o
and Theorem 3 is obtained on the basis of Theorem 2. Since f(()) is equal almost everywhere in (0, 2TT) to the derivative of F«(j), is bounded for 0
Theorem 3 leads to the important
which holds for
Corollary 1. Poisson's integral (1) has almost everywhere on the circle
(1)
< r < 1. It follows from the obvious inequality 0< p' < P and all x ~ 0, that, if u(r, () E h p '
xP
,
< 1 + x P,
then u(r, ()) €
h p " so that the class h p is broadened with decreasing p.
I(, I = 1 limiting values e qual to f«()) along all non tangential paths. An analogous result holds for the more general integral, known as the Poisson
We begin with the representation of harmonic functions by means of a
Poisson-Stieltjes integral. To do this, we need as a lemma the following
Stieltjes integral
Theorem 1 (Helly). Every uniformly bounded infinite family {[(x)! of real 2"
u(r,
e)=2~) P(r,
t-e)drJ.(t),
(7)
functions that is of uniformly bounded variation
2)
in the interval (a, b) contains
a sequence that converges at all except possibly countably many 3) points in (a, b) to a func tion of bounded variation.
where a(t) is a function of bounded variation in the interval (0, 2TT), which can always be extended as a periodic function outside that interval.
1)
Specifically,
Proof. Obviously, it will be sufficient to consider the case in which all the 1) Fatou [1906].
1) The integral (1) can be regarded as the integral (7) with aCt) = f~ f(t)dt. This follows from the properties of a Stieltj es integral.
2) This means that the total variatioqs of all the functions in the interval (a, b) are bounded by the same finite number. 3) Helly' s theorem still holds even without this exception.
386
§2. REPRESENTATION OF HARMONIC FUNCTIONS
IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS
functions f(x) are nOildecreasing in the interval (a, b).
1)
Consider the set of all
By using a diagonalization process, we can select from the family {f(x)1 a
f k (x),
for k = 1, 2, .. "
Let us show that this sequence converges not only at these points but everywhere I'
u(r, g) belong,to hI' For u(r, ()
\'1
00.
nv
)
> -
f(r
ni
) + (v + 1) 1', that is, f(r
nv
) _
00
f
all points for which (2) holds with different
f
v ' we see that
f k (x)
(r, 8) dB
=
21t (Ul (0)
+
U2
(0)),.
(3)
I ~=o = u (0).
Then it follows that
hI'
he Consider the function
''\It u (r, t) dt which depends on the parameter
r, where 0
< r < 1.
Since
2ll
l(lr(R)i~ ~ u(r, t)ldt~M o
as
a sequence of values I'v that approach zero and choosing
U2
0
Conversely, let us suppose that u(r, () €
"
But this is impossible because of the uniform boundedness of the func
tions f(z). By giving
u(r, ~) €
2~.
-+00
2"
+~
Her~ and in what follows, we shall write u (r, ()
k lim fk (r n._l)
k llm fk (r n) -
2"
~ I u (r, B) I dB ~ ~ Ul (r, 8) dB o ~
and n
n
k=I
k=I
converges
I
~ I(lr (B k ) - (lr (B k +1) I ~ ~
everywhere in (a, b) except possibly at a countable set E of points. The limit ing function f(x) is defined and monotonic on the set (a, b) - E. Consequently,
0=B 1
at points of the set E it has left- and right-hand limits. If we define the value
K, 2"
1 \" u(r, B)=2 ~ U(Pk, t) 1t
0
pi/- r2
'9
h -
Pk cos (t - 6)+ r odt.
2' r
388
IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS
If we integrate by parts, we obtain
21CU (r, 6) = a pk (21C)
pi/ _
aQ.b f3 ~ aa + {3b.
Ph"-r'
2pi/ cos 6 -f..-. r·
~
-
t a
Pil ( )
iJt - '.
'2
<x (x)
To see this, note that the function
21t
--+
and {3 are all positive and a + {3 = 1, then
Proof. We note that, if a, b, a
\' a '
If we now take the limit as k we have
389
§2. REPRESENTATION OF HARMONIC FUNC'tIONS
and remember that ap'k (217) = 217U(0) = a(217),
Since 1>(b)
=
0, it follows that
1> (x)
~ 0 for 0
~ b; this proves (5) for a ~ b.
<x
By reversing the roles of a and b, we can prove (5) in an analogous manner for 2"
a ~ b. Let us replace a, (3, a and b in (5) with lip, l/q, a P and b q • Then, for
21CU (r, 6) = a (21C) P (r, 6) _ ~ a (t) aP (r'"t - 6) dt. Inverse integration by parts now yields formula (7) of
§ 1.
a, b, p and q all positive with lip + II q
=
1, we have the inequality
I I q .ab,,;::::-aP+-b -='p q •
This completes the
proof of" the first part of the theorem. The second pact is proved analogously.
(6)
, Now, since the functions Ig (x)! P anJ Ih (x)! q are summable on E, it follows Corollary 1. For a function u(r, e) to belong to the class hI' it is neces from''fM that g(x) h(x) is summable. Furthermore, from (6) we have sary and sufficient that it have a representation in I" < 1 in the form of the dif .
\
ference of two nonnegative harmonic functions. Proof. Suppose that u (r, h) E hI
Then, u (r, h) can be represented in
"1 1
and
M is
391
Theorem 3. For a function u (r, 0) that is harmonic in the disk
finite and independent of n, then
lim ~ In (x) dx
n-+co E
1'1
have a representation in the disk
= EV(x) dx.
(8)
1. As
2",
Let us now turn to the representation of functions that are harmonic in 1 by means of Poisson's integral (1) of
§1.
For example, the function
I~
+~
(r, B) = bo
is regular and bounded in
Then, its real and imaginary parts represent bounded harmonic functions
n bnr sin nB),
n
(bnr cos nB
+ anr
sin nB),
n=1
f«()
I(I
=
1 as (approaches
also has definite limiting
values along nontangential paths almost everywhere on that circle. These limit ing values define a function on
n
I(I < 1. in 1(1 < 1
which belong, in particular, to the class hi' In accordance with §2, these func this circle along nontangential paths. Therefore,
(anr n cos nB -
n=1
V
f«()
tions have definite limiting values almost everywhere on
from Parseval's formula applied to the expansions u (r, B)
Specifically, suppose that a function
1(\
= 1, which we shall denote by
f«()
or by
f(e i8 ) where (= e i8 . This very simple result can now be generalized to the class N (introduced
which constitute the real and imaginary parts of the Taylor expansion ex:>
by R. Nevanlinna) of functions
f«()
u+lv= ~ (an+lbn)~n.
Cf'
Specifically, in this case, we have We can always assume that ¢ unity in the disk 2) A proof can be found, for example, in the book by Zygmund [1959).
1(\
< 1 and that
(C) •
/(q=~ (C)
Ii=O
I)M. Riesz (19271.
that are regular in the disk
can be represented in that disk in the form of the ratio of two bounded functions:
«()
and
lfr«()
(1)
are bounded in absolute value by
1(\ < 1.
Another way of characterizing the class N is on the basis of Nevanlinna's theorem:
394
§3, LIMITING VALUES OF ANALYTIC FUN\jTIONS
IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS
Theorem 1. For a function f«()
i
0 to belong to the class N, it is necessary
Cf'r(Q=
and sufficient that the integral +
2"
~ log If (re i8 ) I dB o
be bounded by some constant M for 0 Proof.
1)
If the function f«()
i
~r (C) = e
< r < 1.
+
rk
2"
o
cm(m + c
(3)
0
2"
mt1
(m t 1
1: d1t ~ log I ~ (re ) I dB= log I em I+ o iO
log I ~k I
is over all the zeros of 0/ «() that lie in 0
0/(0
in 0
< I" < 1
(4)
< I" < r. Since the right-hand mem < r < 1, the right-hand member....
1 in that disk. If we take a sequence of numbers r k in the interval 0
that approaches 1 as k ---+
00,
1'1< 1
to a regular function 0/ «(\Also,
I"
¢ «()'t~!_~s!~gular in proof of the theorem.
I"
< 1;
also
11> «()I
From the representation (1) for the functions f«() r
< 1,
k
logf(Q=
1:
log
ICkl(t::")
and
Suppose that the sequence
r/J(O
~ 1 definite limiting values f(eiB).If we again apply Fatou's
"1;0
The finite limiting values of the function f(t::") €
I.
log
If we apply this inequality to the functions
N along nontangential
As a consequence of Theorem 2, we have the following uniqueness theorem.
< r,
f n (t::")
(9)
and use inequality (8), we ob-
loglln(~)I:o;;;;:; r+!:! M. If we fix
t: "
in the disk
It::" 1< I
\t::"1 < 1. Obviously f(O = [, (t::") - [/t::")
equal in the disk
€ N. If [(01=0 in It::"\ < I, then, by Theorem 2, Iloglf(e i8 )11 is finite almost everywhere on I" = 1; that is, f(e iB ) is finite and nonzero almost everywhere on I" ;0 1. This contradicts the condi tion according to which [(e i8 );o 0 almost everywhere on E (except at points at 00).
In connection with the boundary values of functions in the class N, we also have the following convergence theorem:
and let r approach I, we obtain the inequality '+1 CI M
Theorem 3. ') If two functions [, (t::") and [/0 € N have the same boundary values on a set E o[ positive measure on the ei~cle 1t::"1 ;0 I, they are identically
;0
I"
tain the result that, in such a disk
values.
[2(t::")
we obtain in any disk
I/(~) I ~ 2~ ~ log I/(fe i6) I ffi (~:~:~~ ) dO.
paths, which exist almost everywhere on It: " I ;0 I, will now be called its boundary
;0
< r,
23
\\.""..
This completes the proof of the theorem.
which [, (t::")
converges at all points belonging to a subset
terms o~ the right are nonpositive for where 0 \. r < 1,
have almost everywhere on
lemma to the integral (6), we conclude that jloglf(e i8 )11 is summable on
Proof.
I" I"
(8)
= 1. Then this sequence converges uni formly in the disk 1t::"1 < 1 to a function belonging to the class N.
I definite limiting values along nontangential paths and these limiting
It: "1
I[n (01
E of positive measure of the circle
values are nonzero almost everywhere. But then the function f(t::") has almost everywhere on
+
~ log lin (fe I6 ) IdO ~ M. o
Proof. If we take the real parts in formula (5) and note that the first twO
is an arbitrary function belonging to the class N, then in
its representation (1) the functions
"I =
23
said regarding the integral (6), the right-hand member
are almost everywhere nonzero on
f(O 1= 0
the interval I" < 1. Suppose that there exists a single finite constant M such that for every n and for 0 < r < I,
(7)
Consequently, /logl[(e i8 )!1 is summable on I" = 1. But then, the values of log If(e iB ) 1are finite almost everywhere on "I = I; that is, the values of [(e i8 ) Now, if
Theorem 4.1) Let I[n (t::")1 denote a sequence of functions that are regular in
r-1o
of this inequality is bounded.
397
lin (Q 1:0;;;;:; e'-ICI This inequality shows that the functions disk
I" < I.
f n (t::")
are uniformly bounded inside the
\t::"\ < 1.
Let us suppose now that they do not converge throughout
I"
Then, just as in the proof of Vitali's theorem (§2 of Chapter I), they contain two subsequences I [n'k(t::")l and Ifn"k(t::")!, for k;o 1,2"" that converge in < I to distinct functions that are regular in that disk. By hypothesis, the sequence of
"I
the differences [k (0 converges in
O. It follows from this last p fact and §3 that a function fC() € Hp , for p >0, has, almost everywhere on the circle \(1 = 1, definite limiting values along nontangential paths and these form a limit function fCe iB ). In accordance with Fatou's lemma applied to the integral (1), by letting r approach 1 we conclude that the function fCe iB ) € LP in CO, 211). Theorem 1.
2"
BU~i,~ce IbpCOj:::; 1 in \(1 < 1 and IbpC()\ == 1 on 1(1 = 1, it follows that If~C()1 :::;~COI in .1(1 < 1 and IfpC()1 == Ih/Ol on 1(1 = 1. Consequently, in
< r < 1.
Obviously, all functions that are regular and bounded in P
0 to 211, we obtain
(1)
o is bounded for 0
e from
~ I hp (re i6 ) IP dB ~ ~ I h p (e i6 ) \P dB. o 0
2n
=
Ihp(eit)IPP(r, t-e)dt.
o
§4. Boundary properties of functions in the class Hp
f pC()
1(1 < 1
\hp(reiB)IP~2~ ~
function.
r in 0
403
p
2n
Corollary 1. Every function fC() ~ 0 belonging to the class N has a Blaschke
in
H
where bpC() is its Blaschke function and h pC() .;,. 0 in 1(1 < 1. The function [hpC()]P is regular in 1(1:::; L Consequently, from Poisson's formula applied to
the integral (11), we obtain
class H
IN CLASS
1(1 > 1 -
i6
M
,n.
But since this integral is a nondecreasing function of r in the interval 0 inequality (4) also holds for 0
o < r < 1.
< r:::;
0,
such that, or any set e with mes e < 0 we have Ie IfCe,B)! P dO < l in the interval CO, 217). Sin e IfCreiB)1 - ..... IfCeiB)1 almost everywhere in CO, 217) as r -+ 1, by taking an arbi y sequence of numbers r < 1 that approaches 1 as n -+ we " conclude on the basis of Egorov's theorem that there exists in CO, 217) a set e 1 such that mes e 1 < 0 ac.d IfCr eiB)1 -+ IfCeiB)1 uniformly outside e as n -+ " 1 Therefore, if we deno{e the complement to e 1 with respect to CO, 217) by E, we 00,
Proof. It will be sufficient to assume that fC') ~
o.
If we represent fC') in
accordance with formula (2), we conclude from Theorem I that the function
00.
[hC,)]p/2 € H 2. Consequently, by virtue of Theorem (2) of §2, the real and imag inary parts of this function can be represented in \(1 < 1 ·itt terms of their limit ing values by means of Poisson's integral. But then, the function [h C,)]P /2 itself wilrhave such a representation in "I
= 2~ .~.
lim
< 1:
~ II (r "e iB) IP de = ~ II (e IB)
,,-00 E
2"
hP/2 (re iB)
have
hP/2 (e/ i ) P (r, t -
E
(7)
iB
~ I/(r"e ) IP da= ~ I/(e' B)
"_00 el
In accordance with the Bunjakovskil-Schwarz inequality, this yields
o
2"
per, t - a) dt,
Ia -
o
b IP ~ ( I a I
e1
+ I b I )P =
2"
~ I h (re/ B) IPda ~ ~ 1 h (e lt ) IP dt. o 0
But since
IfC')1
~
jhC')1
in
\(1 < 1
and IfC')
we obtain from (8), for 0< r < 1,
=
IhC,)!
almost everywhere on
l.
Therefore, there exists a
for
(max (21 a I. 21 b I»P
I b IP) ~ 2P (i a IP + I bill),
for arbitrary complex numbers a and b, we have (8)
2"
~ I/(r "e
iB
) -
o
I (e iB) IP da ~ ~ I/(r"e tB ) -
l(e iB )1 P da
E
+ ~ 2"( I/(r"e/BW + I/(e iB) IP) de ~ 2~e +"2 P2e = el
1) F. Riesz [1923].
l
"
= 2 max (j a Ill,
so that
1'\ ~ 1,
IfCr eiB)IP dO
0, then, in the representation (2), we have Hp . Consequently, Poisson's formula (13)
Proof. If fC') €
by the formula
hC/J €
2"
¢ Ct) E L for
t
2"
211:
2"
o
0
.
hP(C)=2~ ~
CO, 21/.), then it easily follows that, for 0 < r < 1, €
Hp ' where p > 0 and if IfCeie)l $ M almost every
where on 1(1 = 1,. then IfC')1 ::; M in 1'1 < 1. Furthermore, if fC') € Hp ' where p >0, and if fCe,e) € Lq, where q> p, then fC,)€ Hq•
f(C)=in~ ~(t)P(r, t-6)dt, is applicable in 1(1 1(1 < 1 and IfCeie)1 to the inequality
~ Ij(re iB ) I d6 ~ ~ I
1. We conclude from inequality (14) that its left-hand member, being indepen < 1 in terms of Poisson's integral, it is necessary and sufficient
1)
1(1
407
p
r in 0 < r < 1, bounded as p ---> 1. Therefore, in accordance with (6) with p = 1, we conclude that this integral approaches zero as p ---> 1. Furthermore, in the
n ....... ooQ
Theorem 3.
H
But in the first of the integrals on the ~ight, the Poisson kernel is, for fixed
is an arbitrary positive number, this means that
lim ~
IN CLASS
h (e it) P(r, t-6)dt P
o
< 1 to the function [hC'W € H l' Since IfC')1 ::;!hC')1 in = jhCe ie )! almost everywhere on 1'\ = 1, this formula leads 2"
If(C)IP~2~~ If(ett)\PP(r, t-6)df.
HI'
Conversely, if fCI;) €
o
H l' then, for 0 < r < p < 1, we have
2"
I \
'!h
Therefore, if IfCeie)1 ::; M almost everywhere on
pI - r S
it
f(re' }=2n ~f(pe )ps_2prcos(t_e)+rsdfo
IfC')1 ::; M in 1'1 < 1.
\'1 = 1,
it (ollows that
On the other hand, if fCe ie ) E U, where q> p, then, by
using Holder's inequality, we have
Therefore, 211
If(re/ B) -
2~ ~ f(l1
it
)P(r, t - 6)dt
If(C) IP ~ 2~
I
211
If(peit)-f(e it )
I
S n
~
pl_r ..
M'
..
•
l
~ 2~ (~
dt·
\ \ I pI - r +21t~ If(e it )IDI-2orcos(t-O)-I-rl-P(r, t-6) I dt.
1) Fihtengol'c [1929].
~
If(e it )
IP PPl9 (r, t -
q-p
6). P-q- (r,t - 6) dt
·211
211:
~~ 2n \~
~
q_p
211
If(e
it
9 )1
P (r, t - 6) dttq .
(~
per, t - 6) dt)-q 211
(14)
= (2n;p/q 1) V. I. Smirnov [1929].
(~If(eit) Iq per, t - 6) dt
r
9
•
~~-__ ~:..::-~~~
~ __ ~_~~_ -~~-:
408
~=~~~
"~~~~~_~~_~~~_-,=-_~o~~p~~
__
~~~_~~~--.-=~~~~~_~~:_~
__
__
~~"~_~~_=_~~ ~_ -=-_~~~~~" =~~ ~~~~~~ _~_=~~~~_~ ~"_"__
_=_~~_~=~~~ ",~~~~~~~~_~~_
'._'--~~--::-~=..:::=-:.-:=-~~="'=--"---~=:-;--=-:::::.:::-~~"=--~:~:~:::.::~:~,':~~~~--=,::::-=~=-,:=;==-~~~,:,,":,,~:-=:=;;:::;-::-~~--~~=~~~~="-,,"..,...~
IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS
§s.
that is,
in
2",
2",
'S
o
I(I
=
q
r, we obtain, for 0
If we rule out the case F«() == 0 (when what we are trying to prove is
> 0 in 1(1 < 1. Therefore, if F«() = Re1q" then 1C1>1 < 17/2 in 1(1 < 1. Now, suppose that 0 < P < 1. If we set u p (r, () = 3t([f«(W) = RP cos pC1>, we have u p (r, () > 0 in 1(I < 1. Therefore
If(c)lq~2~ ~ If(e/t)lqP(r, t-fJ)dt. If we integrate this over the circle
1(1 < 1.
409
FUNCTIONS CONTINUOUS ON A CLOSED DISK
obvious), we have 3t(F«(»
< r < 1,
RP = Ell. (r;
6) & up (r, 0) cos p~ --=:: It
2",
If(re/ e) I dfJ ~ ~ If(e it ) lq dt.
•
cosP2
0
If we integrate this over the circle
rI J
This proves that f«() E H q . This completes the proof ofthe theorem.
2.,
Theorem 5. 1) If the function f«() is regular in the disk '(I < 1 and if f(re ie ) - .... f(e il ) almost everywhere on the circle 1(1 = 1 as r -+ 1 and if
1(1
=
r, for 0
--k- r
< r < 1,
we have
2..
F (rete) I
P dfJ
~
cos"2
Up
~
(r, fJ) dfJ =
p
2ltU
~O) •
cosP2
f(e Ie) E L in (0, 217), then, for Cauchy's formula Consequently, F«(Y E H
.1) P
21
1 (df(reiB) / dO) = ie iB f' (e iB ) almost
of Poisson's integral:
everywhere on k
lrj' (Q =
ix ~leltj' (e
The function
)
P (r, t - a) dt.
(3)
t
g(t) =
lei/j' (e it ) dt
~
~_____
o
(4)
~
is absolutely continuous in (0, 217) and, in accordance with
C~y,S theorem,
g (217) = O. Integrating (3) by parts, we have 2"
6) dt,
2"
1(1
=iJ~ ~
Proof. Since f(O can be represented in
< 1,
If(e 1o ) - f(e 1o ,) I ~KI on the circle 1(1
where c (r) depends only on r. But c (r), being the difference between two func
1(1 < 1,
is itself a harmonic function. Since
!i.e (r) = die (r) drs
we easily obtain the result that c (r)
=
(0
I(I < 1
by means
E H l' It follows on the
basis of Theorem 1 that f(O is absolutely continuous on the circle
1(1
=
1.
+ 1-r dedr(r) •
If'(q I ~ Proof. :For a
0< a
< 1.
=
0
< a.~ 1,
11 !:.\1 ,., r = Iq,
where M is a finite constant, be satisfied in 1(1 =
(6)
(7)
< 1.
1, the theorem is obvious. Let us consider the case when
Suppose that f(O is continuous on \(\ ~ 1 and that inequality (6) is
satisfied on 1(1 = 1. Since f(O can be represented in with Cauchy's formula
alog r + b, where a and b are constants.
Since c(r) is continuous at r = 0, it follows that a
a- a' [",
1, it is necessary and sufficient that the inequality
=
o
tions that are harmonic in
in accordance with formula
imaginary parts of the function i(f' (0 can be represented in
f(Q=~ g(t)P(r, t-a)dt+c(r),
,
1(1 < 1
(1), we find, just as at the beginning of the proof of Theorem 1, that the real and
g(t)P(r, t-a)dt.
211
Thus, in
ous on the closed disk 1(1 ~ 1 and if it is a function of bounded variation on the circle 1(1 = 1 (that is, if the real and imaginary parts of the function f(O are functions 0 f bounded variation on that circle), then f(O is absolutely continuous on the circle 1(1 = 1.
Theorem 3 (Hardy-Littlewood). For a function f(O that is regular in the open disk 1(1 < 1 to be continuous on the closed disk 1(1 ~ 1 and to satisfy a Lipschitz condition
that is,
Therefore, in
1. This completes the proof of the theorem.
of a Poisson-Stieitjes integral and, consequently, f'
l~f' (q = ~ g(t) iJP (r,,,,~ -
iJftei8)
r
=
Theorem 2. If a function f«() is regular in the open disk I( I < 1 and continu
, it
"I
211
f(q=~ \ f(elt)e lt 21t ~ elt - t dt,
0 and hence c(r) == const = c.
1(1 < 1, it follows that, for I( I < 1,
1(1 < 1
in accordance
412
IX. BOUNDARY PROBLEMS FOR ANALYTIC FUNCTIONS
I' (Q =
21t
§5. FUNCTIONS CONTINUOUS ON A CLOSED DISK I
21t
~ C f (e it ) eit dt _ ~ C f(e it ) - f(e la ) it dt 21t .~ (e lt _ C):i - 21t.~ (eft no e , o 0
If(eiB) -
h
I
+~
where'; = re it . Consequently,
, I e If(eit)-f(e ia ) I k
If(C)I~21t~ o
"" Ie If(eil (l
1!(re i9 ) _
+ r)/Z, so that
!(r'ei~ I ~ M
IT'T
(1- t)" a
r' - r>
1-
r',
converges, then f(') is bounded in the open disk then
< M (1- r')" ~ ~ Ir' a
a
disk
1'1 < 1
If a function f(')
=
r I".
1'1 .: ; 1
closed disk
and satis
0< ~< 1,
a' I",
where (, = re iB ). Consequently,
"\
I
I!(QI~-;- j
-11
" 2\
A (t)
A (t)
II-reitrdf+Mh
where MI is \llso a constant. If we replace the absolute value in the denominator of the integral on the right with its imaginary part and use the inequality sin t
Proof. Since f(O can be represented in the open disk
1'1 < 1
in accordance
"
..
+
I \ eit C f(~)=21t ~ u(t) eit_cdt+iC,
2"
2u (t) eit dt _ 21t ~ (e it - C)s -
1.- \ 1C
~
u (6) it dt (elf - c)~ e •
u (t) -
Consequently, just as in the proof of Theorem 3, we obtain the inequality
If' (Q l~
11
~\I--'"
f(')
Theorem 7. I)If a function f(')
2..
1.- \
""2
1 !(~)1&2 -= 1t ~\ ~dt+M"':'::1.r 810 t -== r \~ ~df+M t h
(14)
which proves the boundedness of
it follows that
>
(Z/77)(, we obtain
"2
2..
I)Privalov [1919].
""2 2 \
I!(QI~-;- ~ II_reitldt+M~-;- ~
with the Schwarz formula
f ... -
dt
.
IU(f+6)-u(6)III_relf l +M,
where M is a constant. Therefore, in accordance with (15), we have
(13)
then f(') satisfies a complex Lipschitz condition on the
' (T) _
We transform this formula to the form
1t
I u(6)- u(a') I ~Kla -
1'1 = 1, 1'1 .: ; 1.
< 1.
1\ elt+C !(q=21t ~ (u(t)-u(9)) ..If_r df+u(9)+iC
fies the Lipschitz condition
on the circle
I"
2"
u(r, e) + iv(r, e) is regular'in the open
and if u (r, e) is continuous on the clos ed disk
1'1 < 1.
Proof. Formula (14) holds in
the theorem. 1)
(15)
where the function A(t) is nondecreasing and is such that the integral fc{(A(t)!t)dt
In both cases, we have obtained (11) for suitable K. This completes the proof of Theorem 5.
and satisfies on the
lu(e) - u(e')1 .::; A(je - e'l),
" =Mjr' -r)".
T
On the other hand, if
I' I ~ 1
1 the condition
T'
1!(re'9 ) -
415
ance with Theorems 3 and 4, it follows that equation (14) holds. This completes Theorem 6. If a function f(O
~~
(1
1'1
FUNCTIONS CONTINUOUS ON A CLOSED DISK
the proof of the theorem.
0,
J
t~h
1.:1 .$ 1 and is
1;
a
\ s(t+h)-s(1:)d _1
=
3) dw(eit)/dt
points, we have in (a, b) by Fatou's lemma t t lim \ S (c h S (t) d't ~ \ S' ('t) d't, a
of Chapter IX and the lemma just
curve C, then thg following hold: (2)
On the other hand, since s(t) is obviously a nondecreasing function, it is
+ k-
§5
w (.:) is regular in 1.:1 < 1 and maps that disk univalently onto a finite domain B that is bounded by a rectifiable closed Jordan Theorem 1. If a function z
a
= ~[
of Chapter II, the function
proved we immediately.obtain
t
a
Let z = w (0 denote the
2.". Since C is rectifiable, the function w(e it ) is of bounded variation in (0, 2.,,).
S(t) 0, then on
'(I = 1
there are closed sets
Gt,k), for k = 1, 2, " ' , such that Gt) C E sand mes G k ) --+ mes E s' Let us
l
k
k
t
for k=l, 2,"',
denote the corresponding set on C. For any (> 0, there exists on the circle
define GS=U:=lGt ). Then, GscEsand mesGs>mesG
1(1 = 1
so that mes G s = mes Er • Consequently, the set N s = E s - G s is a set of measure
an open set 0 s such that Esc 0 sand mes Os 0, a harmonic function iri the closed disk 1(1 ~ 1.
We assert that there exists a finite constant K such that lu «(, ,)1 < K for all (
which is, for arbitrary fixed ,
in the closed disk 1(1 ~ 1 and all , in the interval 0
00.
Furthermore,
if necessary, ·arrange to have
1 at which lu «(n'
'n)1 = mn
'(I = 1
at the point (0' This con
1(1 =
denote the angle between the tangent to C at the point a,=
weal
I( \=
1 at the point a. Then, for given f
5> 0 such that, for
Irl
1 and let ¢ and the tangent
> 0, there exist "I > 0 and
< 5, we have
't) -lfll
=
1< 1 in accordance with Poisson's
2"
arc
the point a mentioned above, we have the limit relation u «(, ,) --> arg w ' «() as
have, for 0
formula
=2~ "'t" ~ V (e lt ,
1(\ < 1. Consequently, there exists an "II > 0 such that the inequality '(-al 1.
Then w"(,) € H p '
Proof. From the condition of the theorem, we find, by using Holder's inequality $
where k 1 and k 2 are constants, hold on the circle
\'1
=
I(j (s) -
1.
B(s') I = ~ B' (s) ds':;;; $'
S
S
0 B' (s)P dstP 0 dS)I/q ~M Is $'
s' liN,
s'
I
Proof. By Theorem 4, we have
arg w' (Q=B (Q - arg ~- ; , and, by Theorem 5, we have w '(,) € Hp for all p and formula (3'), we have, for ()'
> o.
(8)
where lip + llq
=
1. From this we conclude, by Theorem 6 (With a= 1Iq), that
w'(,) is a continuous function on the closed disk dO (s) = dO (s) - dO ds
< (),
8'
\ I
I,
w' (e ia ) 12 dB ~ dB ~ k 1 I B
B' jI/2,
arg w' (e i6 ') I ~ IB(s) B(s') ~ k 2 1s
1+\ B
- s' /a
~ (argw' (e I6 )+B + ;) E!.p.
B' I
+I B-
But, by Theorem 3 of §4, Chapter IX, and Theorem 2 of § 1, Chapter IX, we have
B' 1< k 2 1 B- B' la/2.
almost everywhere on the circle d
It follows on the basis of Theorem 5 of
§ 5,
8
I = ~ I w' (e i6 ) IdB ~ ks·1 B6'
and, consequently,
B' I,
k s = max I w' (Q Itl=1
I.
1'1 =
,e ,,'e
Ie
1000' (e I6 ) 00' (ei8)
,
,'e ») € LP. It now follows from the representation of the
so that ~ (e' w (e' )/ w (e' function
I
log 00' (e i6 ) dO
Chapter IX, that log w '(,) and
hence w'(,) are continuous functions in the closed disk \,' S 1. But then, s'
.1 w' (e i6 ) I,
we conclude that d() (s)1 d() € LP; that is,
8'
so that
Is -
Since
8
6
Is- s'l = ~ I w' (e i6 ) IdB ~
I arg w' (e i6 ) -
1'1 S 1.
On the basis of this fact
'W '(,)/ w '(,) by means of Poisson's integral that
1) Smirnov [1932].
428
X. FUNCTIONS ANALYTIC INSIDE A RECTIFIABLE CONTOUR
ffi.
(COO' (I;)) 00' (C)
'§2.
PRlVALOV'S UNIQUENESS THEORE."'M
429
Define fn (z) at the point z 0 € E as the maximum value of Ifn (z) I in the sector s. Obviously, this sequence converges to zero on E. In accordance with
E hp' '
%0
Then, by Theorem 4 of
§ 2,
Chapter IX, (w" «(V w I «() € Hp' Consequently,
w "«() € H p • This completes the proof of the theorem.
§ 2.
Egorov's theorem, the set E contains a closed subset P of positive measure on
co~verges uniformly to zero. The complement of P
Ifn (z)\
which the sequence
wi th respect to the disk
Iz I =
1 is an open set and hence consists of finitely or
countably many disjoint arcs. We replace those arcs that are less than a semi
Privalov's uniqueness theorem
Let us agree on some terminology. Suppose that B is a domain bounded by a Jordan curve C and that z 0 is a point on C at which there is a unique tangent to C and in a neighborhood of which the curve C lies on both sides of the normal. We shall refer to a continuous curve l contained in B and ending at point z 0 as a nontangential path if that part of l in a neighborhood of z 0 lies inside some angle of magnitude less than TT with vertex at the point z 0 and with bisector coin
circle with the sides of the right angles intercepting them from inside the disk
Iz I < 1
Iz I =
in such a way.that the sides of the angles intersect the circle
1 at
an angle TTl 4. On the other hand, we replace an arc greater than a semicircle with the two radii that pass through its endpoints. As a result, we obtain in place of the circle
Iz 1= 1
a rectifiable closed Jordan curve C 1 bounding the domain B 1
and contained in the disk
f (z)
Iz I < 1.
Uere, the set P lies entirely on the boundary
C l' Tbe function
ary of the domain B, we shall usually mean only points at which there are tan
points of the set P. Let us now map the domain B 1 onto the disk
gents to C possessing the property just described. Furthermore, if a function
w «() denote the inverse of the function executing this mapping. Correspond irig to the set P is a measurable set PI of positive measure on the circle 1(1 :51,
f(z) that is regular in B approaches a value a as z approaches a point z 0 on C along every nontangential path contain~d in C, we shall say simply that f(z) assumes (or has) the value a along nontangential paths. We turn now to the investigation of functions that are regular in a domain'
z
is continuous in
lJ
ciding with the inner normal to C. When we speak of a set of points on the bound
if we assign to it the value 0 at
I( I < 1.
Let
=
The function f 1«()
=
f(w «()) is regular in the open disk
1(' < 1,
continuous on
tbe closed disk \(1 S 1, and equal to zero at points of the set P l' Then, in accordance witb
§ 3 of
Chapter IX,
f 1 «() ==
0; that is,
f (z) == 0
in Iz 1
< 1.
bounded by a rectifiable Jordan curve. We begin with the exceedingly important
Thus, the theorem is proved for the case of a circle.
uniqueness theorem of Privalov.l)
Let B denote an arbitrary domain defined by a rectifiable closed Jordan curve
Theorem 1. Let f(z) denote a function that is regular in a domain B bounded
C. And let us map it onto the disk
'(I
by a rectifiable closed Jordan curoe C. Suppose that f(z) assumes the value 0
of positive measure of
along tangential paths on some set E of positive measure on C. Then f(z) == 0
transformed into a function
in B. Proof. Let us consider first the case in which B is the disk
Iz 1< 1.
About
every point z 0 € E, let us construct a right angle with vertex at z 0' with rays directed into
\z I < 1,
and with bisector directed along the radius to
us cut from this angle sectors
s,
for n
=
z=
O. Let
1, 2, ... , of radii lin. Consider a
sequence of functions fn(z), for n = 1, 2, ... , that are measurable
2)
on E.
=
'(I < 1.
We denote by E I the measurable set
1 into which E is mapped. The function
f 1 «()
that is regular in the disk
1(1
O. Let h = h (.,,) > 0 denote a number
Now, suppose that we are given ." such that, for \s - sol
< h,
that tangent. H we can prove this assertion about the difference· (4), then, almost
Z'(S)--'-z~
everywhere on C and in particular at points at which f(z') is finite, the existence
s-so
of one of the integrals (4) will imply the existence of the corresponding limit for Let
the other integral.
10
denote an arbitrary number in the interval 0
This assertion is easily proved if fez') == 1 on C. Let z ~ = z '(s 0) denote an arbitrary
C\
in log (z ' -z) for one circuit around the curve C. The imaginary part of this from the point z to the point z'. But (log (z / - z))c does not- change if we re
f), Z /(s 0 + (0)) on C with a semicircle y with the same endpoints located on the other side of C from z. On the other hand, the increment place the arc (z '(s
0)
± 1Ti
(z') -
~
rp (a) da (1 +m) a-elei,!,
Therefore, we can write (5) as follows:
O. Obviously, the value of the subtrahend in (4) is equal to this last ,
can be made arbitrarily close to
10,
(f (z· (s» - I (z~» dZ~~S) ds -------:-.-----.:.:...:/;.,-.,....,-, C (S-8 (e'Io+a)_eie (10+'1')
\
I (z~) dz'= ) z'-z
CI t;
F (z,
increment de creased by the increment in (log (z ' - z ~))'Y' which, for sufficiently small
z -z
,
dz =
We have
0 -
in log (z ' - z~) for one circuit around this new path by an amount that approaches 10 - ....
1
< 10 < h.
If we now set s-sO=a, oe- i1o =m=m(a), (f(z/(s)-f(z~))(dz'(s)/ds)x e-(~o = ¢(a), where Iml 0, the class of all functions f(z) -t 0 that
~ rf(z) fPds cr 1 +1 2£ .
(11)
and furthermore,
a" (0) We have
Since the function an (X (z)) is continuous on
{g~
)
For sufficiently large n, we therefore have
g(B)dO 0, there exists an r in 0
~ I P(z) -j(rx (z» V
C
~ \ F(z) - p(z) lilds= SI F('t) 19ds+ ~ Ip(z) 19 ds c
C
C
9ds+
1
II
n
~ I Ck 19
C
k=O
ICkI9+
k=O
= 0,
(12)
F(z)-Pn(z)=_l \ F(z')-Pn(z')dz' 21':1 ~
~
I Ck-
z'-Z
.
H E is a closed set in the domain B and if 0 is the distance from E to C, then,
n
II
-2m(~ CkCk)=~IF(z)19ds- ~
C
k=O
C
c k> for k = 0,
if condition (8) is satisfied. Now, by Cauchy's formula, we have in B
C
- 2m (~F(Z)p (z)ds )=~ IF(z)
=
it follows that, for these polynomials Pn(z),
CkI 9.
for z € E,
k=O
IF (z) -
It follows that the minimum for the integral
~ I F(z) -p (Z) 19 dC
Pn (z) I ~
2:e ~ IF (z') -
Pn (z') 1 ds
~
(9)
ice -V~~"-IF-(-z'-)-Pn-(-z'-)/-9-d-s-.l-e-ng-th--'-C.
C
yields a polynomial p(z) determined in accordance with formula (8') with c ~ = c k. Here, this minimum is equal to
~ IF (z)
II
9 1 ds
-
From this .we conclude in accordance with (12) that pn(Z)
--+
F(z) uniformly on E.
Since Pn(z) is a segment of the Fourier series (11), this proves the theorem.
kJ;;{J I Ck 19.
(10)
In particular, if we apply Theorem 6 to the function
f (z) =
.JX '(z) and keep
in mind the fact that then For the system of all polynomials in z to be complete, it is obviously necessary and sufficient that the difference (10) approach 0 as n function
--+
00
for an arbitrary
F (z) € E 2' that is, it is necessary and sufficient that equation
211:
cn
= SV X' (z') Kn (z') ds = ~ K n (Ill (e IS )) -v-;J(ei6 ) dO C
0
(8) hold.
= 21tKn (0) y-;;;-' (0),
This completes the proof of the theorem. we obtain
By combining Theorem 4 with Theorem 3, we immediately obtain
Theorem 5. For an arbitrary function fez) € E 2 to satisfy the closedness formula, it is necessary and sufficient that C satisfy condition (8).
ex>
(~ Kk (0) Kk X' (z) = 21t k=~
We now have
Theorem 6. When condition (8) is satisfied, an arbitrary function F(z) € E 2 00
k=O
ckKk (z),
(Z»)2 ,
~ I Kk (0) I~
k=O
can be expanded in B in a Fourier series
F (z) = ~
Theorem 7. For curves C satisfying condition (8), we have in B the formula
Ck
= SF (z') K k (z') ds, C
(11)
where the convergence of the series in the numerator is uniform inside B.
(13)
§ 1.
GLUING THEOREMS
455
B (2) in such a way that the points x and g(x) in the interval l are mapped into
the same point w of that interval (see Figure 18). Let us now map the univalent 2
CHAPTER XI
""
domain made up of the images G(l) and G(2) of the domains B(l) and B(2) under these mappings and of the interval l uni
I
SOME SUPPLEMENTARY INFORMATION
-1
ICC
•
'I
I
~1
valentlyontothelune B(l)UB(2)Ulin the w'-plane in such a way that the points
§ 1.
± 1 are mapped into themselves. Under this mapping, the domains G (l) and G(2)
Gluing theorems
Gluing theorems establish the existence of analytic functions that obey
are mapped into domains adjacent to some
certain relationships on the boundarie s of certain domains. The nature of the se
smooth curve
relationships is clear from the statements of the theorems we now present.
Al
that is analytic at all
interior points (see Figure 19). H a func
Figure 17
mapping o( the interval l: - 1':; x ~ 1 into itsel(, mapping each endpoint onto
tion w' = !(w) provides this mapping, then the function w' = !(gk(z)) = gk +2(z)' k = 1, 2, will provide mappings of the
itsel(. Suppose also that, as a (unction o( a complex argument, g (z) is regular
domains B(1) and B(2) onto these same adjacent domains. In accordance with
and univalent in a su((iciently narrow circular lune with vertices at the points
the symmetry principle, the functions w' = gk + 2(Z), for
Theorem L Suppose that a (unction x '= g(x) sets up a continuous one-to-one
k = 1, 2, also map the
± 1 and containing the interval 1. Under these conditions, there exist two (unc tions w = (l(z) and w = !2(z) that map the semidisks B 1 : Izi < 1, ~(z) > 0, and B 2:
Izi < 1,
(rom the disk
< 0, Iwl < 1
:25(z)
respectively onto two disjoint domains G 1 and G 2 obtained by making it a smooth cut A that is analytic at all interior
~1~ ~J
points.1) Furthermore, the mappings are such that, on l, i\:;~
/1 (X) =/2 (g(x», that is, such that cmy two points x, g(x) € l that lie on the boundaries o( the domains
-1
Figure 18
Figure 19
Bland B 2 are mapped into the same point A.
Proof. It follows from the conditions of the theorem that the function z g(z) is regular and univalent in a sufficiently narrow segment
the interval l and the circular arc contained in ~(z) ment, ~(G(z)) ~
o.
8(1)
1=
bounded by
> 0 and that, on this seg
We can take the segment B(1) in such a way that the values
"double segments" B(3) and B(4) bounded by the interval 1 and the circular arcs inclined to l at the angle rr/2 n -
1
in an analogous manner. Let G(3) and
G(4) denote the images under these mappings and then let us map the domain
UG(4) UAl
U8(4) U1 in
of the interior angles at the vertices are equal to IT/2 n , where n is an integer.
GO)
Let B(2) denote the segment symmetric to B(l) about the x-axis (see Figure 17).
such a way that the points ± 1 are mapped into themselves. Under this mapping,
In BO) we consider the function w w
=
=
g 1 (z) = g(z), and in B (2) the function
g 2(z) '" z. For arbitrary x E l, these functions map the domains B( 1) and 1) Thar is, such mar any inrerior arc of ir is an analyric curve.
454
univalently onto the lune B(3)
the domains GO) and
the w "-plane in
G(4) are mapped into domains that are taogent along some
smooth curve A2 that is analytic at all interior points. By applying the symmetry principle n - 1 times as indicated, we obtain the mappings mentioned in the
------"----·"-"-_"
456
~ _ _,
..;...,,>.....___....,...:·...
.u,·..................-_.......n - ..... ~ ...-=-~
~=O;;
...
...==;:.;""';.;;;.-;..:'""i';'.Zii,~
. '".;......:·.:.,·;·:~~::.~ 0 and B 2 : Iz I < 1,
~(z)
00,
We now set
"ll
CPn•• (z)=!.;;'Cf.(z))=z+ ... ,
v?""lJ'offi"~'!'>&. __ ~ 'IJZt:L~';;~';WS~~:~'t;.*,~i!n?.m:e,"§J!i.ib.o;:;,'i:lHr"",-,~;§~,""",r;>,~"'~~Rir&'--;";f~~~~\1i""ja
§ 3·
XI. SOME SUPPLEMENTARY INFORMATION
m
in a given n-connected domain B with non degenerate boundary continua, that are given finite point a € B. Then the quantity
1f'(a)1
=
°
§ 5 o[ Chapter
I
VI).
When conditions (2) are satisfied, we have by virtue of the extremality of the function (o(z)
That the extremal problem posed in the theorem has a solution is obvious.
> 0,
the point z
Since I~=lw/(z) = 1 so that the function (z~, "', z~) can be expressed as a linear combination of the functions F~(z ~ , ... , z ~), k = 1, ... , n - 1, the nth of conditions (2) is a consequence of the remaining conditions and we can drop it.
K I , " " Kn : K = U nk -_ I Kk •
Let [o(z) denote one of the extremal functions. Since I[~ (a)1
=
a log II; (a) I-log I/~ (a) I =
is a simple zero of the function [o(z).
m
~
(g(a, Zk) -
m.
at' " ' , zm in the domain B. (Of course, it is not assumed that these are all the
= k=1 ~ (g(Zk'
zeros of the function [o(z) in B.) Consider the Green's function g(z, () for the domain B and the harmonic
I=
log 1/0 (z) I
+ u (Z),
(1)
m
where the Zk belong to B, k
=
1, ...
1, " ' , n - 1, is minimized when zi.
where
the matrix (Zk U
11~:1II,
(z) = ~ (g(z, Zk) - g(z, Zk», k=1
where, in turn, z~ € B for k
=
in B. Also, for [I(z) to be single-valued in B, it is
and satisfy conditions (2) for 1 =
Zk for k
=
1, " ' , m. But then, the rank of
k=l, "', 2m,
1=0,1, ... , n-l,
(4)
= Xk, k = 1, ... ,2m, does not exceed n - 1. This is true because if the rank of this matrix were equal to n (which can be the case only when 2m > n - 0,
then the functions
g/
=
F /(z ~ , ••. , z~), for 1 = 0, •.. , n - 1, would map a
neighborhood of the point
m
(z) = ~ (h (z, zk) -
h (z, Zk»
k=1
(Xl' •••
,x2m) in 2m-dimensional Euclidean space
onto an open set in n-dimensional space containing the point (0, 0, ... , 0). The intersection of this set with the go-axis (g\
be single-valued.
=
g2
= ... = gn-\ = 0) is an open
segment containing the point (0, 0, ... , 0). C.onsequently, the function
Since, in accordance with (10 'H) of §6 of Chapter VI, 21tlool (Zk),
(h (z, Zk»Kl
=-
where w/(z) is the harmonic measure of the curve
21tlooz (zie), 1= 1, ... , n,
K/ with respect to the domain
B, it follows that single-valuedness of v(z) in B leads to the n conditions on 1) AbHors [1947].
=
,m,
with xi.
1, ... ,m.
necessary that the conjugate function
=-
(3)
= xi. + ix~ +k, zk = Xk + iX m +k)
m
(h (z, Zk»Kz
~ 0.
k=1
log 111 (z)
'lI
g(Zk, a»
Po (z;, ... , z~)= ~ g(zi" a),
-gn, defined by
t d = dn dn .i.J k dn dn s
is a sufficiently small positive number,
f
with the same
dv(z)=-
on
U(Z)={,,(
l.to (a) I+ u (a).
Furthermore, if we denote by Kf,l,l a curve in
= J\" t. t
(11)
Turning now to the function (6), let us suppose that, for some q
f0
Then, by virtue of the extremaliry of the function fo(z), we have u(a) ~ O. By using Green's formula (as derived above), we see that this leads to the inequality
(v (z»K
[ =1, ... , n - 1.
exists a subset E of K of positive measute such that U(z) ~ - q for z E E. Then,
and let z approac h a, we obtain
'= f
0) equations
for u(z) we may take the function defined by the boundary values
1.= log Iz-a 1+
log Is (z) z-a
by the equation WI(Z)
ds=O,
K '0. t. I
is a fixed small positive number. This is equivalent to satisfaction of
(9)
(8), If 2(z)1 S 1 in B. Thus, f2(z) € !JJl. If we rewr ite (7) ill the form
'=
du dn
O. (As always, the normal n is directed inside the
domain B.) It follows. that U(z)
by the equation WI(Z)
\
J
[=1, ... , n-1.
is a sufficien dy small positive
f
the n - 1 limiting (as
2~
+ 2a
k=1
number), we obtain Green's formula for the domain
U(Z)+U(Z)=
drot (z)
u(z)----cTrZ ds
UK., t, k
Ka where K f is the set of curves g(z', z)
~
473
0 }
,
(12)
k=1
[=1, ... , n-1, and, by virtue of the choice of the Ak' for k
=
0, 1, •.. , n - 1, they must be
satisfied. But conditions (12) constitute a system of n - 1 homogeneous linear equations with n unknowns Aft> k
'=
0, 1, ... , n - 1. Therefore, this system has
nonzero solutions. Let us now denote by AIt, k
'=
0, ... , n - 1, any nonzero sys
tem of solutions and let us construct from them the function u(z) in (10). We then obtain an inequality in which the left-hand member includes the factor maybe of either sign. This leads to the condition
f,
which
,-, ==
-----,~-
474
=
"'"
~:
--,-._~~~~?~~~~~.~~-b~~~~~
§3. EXTREMUM FOR BOUNDED FUNCTIO~
XI. SOME SUPPLEMENTARY INFORMATION
Now, the conditions under which
II-I
-.A } (dg(z, a»)S ds+ ~ A' } drok. dg(z, a) ds=O. ~ k_ k=1
dn
o
dn
...'t""~~~~-~~~~:W~'~~F--"'1:;M~~~~~~~~~~tw:",~~~~~
(13)
dn
f o(z)
is single-valued in B take the form
I
m
~ (~A0 dg(z,a)+A dn I
dUl1
dn
+ . . . +A
II-I
where the VI are integers. Obviously, VI over 1 = 1, .•.
dUl Il- 1 )SdS=0. dn .
,n,
2:
1, for 1 = 1, ... , n. H we sum (14)
II
m+l= ~ 'Il~n. ~=l
II-I
dg +
~
dUlk
~ Ak dn =0.
-"0 dn
But m ~ n - 1. Consequently, m = n - 1 and VI = 1 for 1= 1, ... , n.
k=1
Summarizing what we have said, we conclude that the function
Since, in addition, everywhere on K we have
+~ ~
f o(d
has
exactly n zeros in the domain B and that it is equal to unity in modulus on K.
II-I
dg -A O ds-
(14)
we obtain
It follows that, almost everywhere on E.
,
wl(Zk)='Il, [=1, .... n,
k=1
H we multiply equation (13) by -A o and equations (12) by Al (for each 1 =
~
!
21t(arg!0(z))K/=wl(a)+
1, '" ,n - 1) and then add all the resulting equations, we obtain
475
Then, by evaluating (arg (f o(z) - wo» K, we prove that the function f o(d - w 0
dUlk AkdS=O.
has, for each w 0 in the ,disk Iwol
k=1
O. Let us denote the extremal functions by f (z, p), thus
= (. F(z,
a), where
kl
f(a)f(z~) satisfies the conditions of Theorem
=
1.
pointing out their dependence on p. Then, we can easily see that between suitable extremal functions we have the relationship fez, pq)
=
1, and conversely,
a. Obviously, equality holds only for the function (16). The second part of
the theorem follows immediately from the first. This completes the proof of the theorem.
~;
~
Of:
Lemma. If the function fez) is regular in the annulus q ~ [zl ~ 1 except possibly at a single simple pole on the segment - 1 < z < - q and if !f(z)1 ~ 1 on
the circles Iz I = 1 and Iz I = q, then If(z) I ~ 1 on the open segment q < z < 1 with equality holding only in the case f(z):= const.
Izol < r 2
1
log~
log
(1)
10 ra
raM~ gn;
rl' r2'
M1, and M2, and where A is real. Since this last
function is regular in the annulus only for the exceptional values of Izol at which A is an integer, it is only in these cases that inequality (1) is sharp. The
problem arises of finding a sharp inequality in the remaining cases. We shall now investigate this question.
I)
We note first of all that, by means of simple transformations, we can reduce the problem posed to the case when rl
=
q,
r2 =
1, M1
=
p, M2
=
1. Hence, we
can consider functions f(z) that are regular in the closed annulus q ~ Iz I ~ 1 and satisfy the inequality If(z)1 ~ 100 the circle
This function also satisfies the conditions of the lemma and it assumes real
< Iz I < 1
< 1. Let us now consider the case in < z < - q. Since /F(-l)! ~ 1 and IF(-q)1 ~ the interval -1 < z < - q every real value of
=
jzl =
1 and the inequality If(z)1 ~ p
q. The question then deals with the maximum value of
If{zo)1
which F(z) has a simple pole on - 1 1, it follows that F(z) assumes on
absolute value greater than 1. On the other hand, it easily follows from Cauchy's
F(z) assumes every value w such that Iw I > 1 at exactly one point of the annulus q < Iz! < 1. Consequently, IF(z)! must not exceed 1 anywhere on the interval q < z < 1 and, in particular, F(zo) ~ 1. H equality holds here, then F'(zo) = 0 and a neighbor hood of the point z 0 would be mapped by w = F(z) into a multiple-sheeted neighborhood of the point w = 1 or of the point w = - 1. But then, F(z) would theorem on the number of zeros and poles that the function w
assume certain values w such that
Izi < 1.
Robinson [1943].
IwI > 1 at
=
certain points of the annulus q
t:>
I:!
.......
~
J
'-'
+ .. I:! .......
ap+t:
I I:!
I:!
'i'3
--.
,....,...
I:!
ap+t
~
'-'
>t:>
...
>t:>'"
J... J'+.. ... >t:>
..1 .......
m=np+q,
.J
where the unfilled-in spaces are occupied mostly by zeros. We now put the
>t:>
columns numbered n, 2n,"', 2np into the 1st, 2nd, ..• , pth positions (without
'-' I:!
..
+
changing the order of the remaining columns) and we put the rows numbered n + 1,
~
'lfi ....................................... ': ...........
--;:
~ I:!
>t:>
~
..
.......
I:! .......
'-'
:lI:!
>t:>
.......
..:;
~
'-'
>t:>
>t:>
'"
~
'iI:!
>t:>'"
,..... ..:;
'-' I:!
>t:>
.......
J
'-' I:!
>t:>
J
.. .. ..:;
,...., ....... '-' >t:>
--;: ..... '-'
1
>t:>
I:!
.......
,...., ~ J '" j
..
,....,...
~
~
>t:>
....... '-'
..:;
.
.................. -
'" ~
'+I:!
>t:>
a p+l
:\
~
~ I:!
...
.......
~
I:!
>t:>
~
.. ..
.......
~ .Q
a, ... >J,
3 ...
... + '-'
--.
.......I:!
I
'-'
I:!
'+I:!
>t:>
~
>t:>
~
,....,
,...., ~
>t:>'"
'tJ
..:;
'-'
.
""
>t:>
':J'
+ 1, ... , n(p - 1) + 1 into 21ld, 3td, ... , pth positions. Then we arrive at a determinant
that is equal to the determinant Ap multiplied by a determinant in which, for every nonzero
q, the qth row from the bottom consists only of zeros, so that the determinant is equal to zero. On the other hand, if q = 0, this last determinant will have the same form as (16) with p replaced by p - 1. Repeating the same operation, we show after p steps that the determi nant (16) is equal to ± A~. This completes the proof of (4). Therefore,
-
D* = lim
m-+ 00
"!.li~ II A~
I
- n2J1..2r
I = plim ..... oo
II
--- --
I A:p I = plim .....oo
n 2J1..2ri'A1n
II
I Ap In =
'!rn II D,
which is equivalent to (6). This completes the proof of the theorem. We turn now to some applications of this theorem.
Theorem 2. For functions that are regular in an infinite domain B with
-----..-......
§6. ON p. VALENT FUNCTIONS
XI. SOME SUPPLEMENTARY INFORMATION
486
487
boundary K consisting of a finite number of closed Jordan curves and having the
arbitr~rily dose to d(K). This proves the sharpness of the inequality mentioned
expansion f(z) = I:=1akl zk, in a neighborhood of z =
in the theorem. This completes the proof of the theorem.
00,
the inequality D :5 d(K)
Theorem 3. For any complex numbers a and b, an arbitrary positive integer
given by Theorem 3 of §2, Chapter VII, is sharp. Here, Dis defined as in Theorem 1 and d(K) is the transfinite diameter of the set K.
n, and a con tinuum E such that d(E) < via
Proof. We shall go through the proof only for the case in which the curves
p n(z)
=
zn +
C 1 zn-1
-
b 1/4, an arbitrary polynomial
+ ... + C n assumes either the value a or the value b some
mentioned in the theorem are all analytic. Then, if we denote by g(z, (0) the
where in the complement of E with respect to the z-plane. The number
Green's function for the domain B, we conclude that this function is harmonic on
\II a -
K. Therefore, for sufficiently small 0> 0, the equation g(z, (0)
on E.
= -
0 define s a
set of closed analytic Jordan curves approximating the curves in K and defining domains
8.
B8 containing
If g8(Z, (0)
is the Green's function for the domain
B 8, if ;Vn.8(Z), for n = 0,1,'" are the Cheby~ev polynomials for the boundary K8 of the domain B 8 which have all their Zeros on K8, if ;;;n. 8 = max z E K 8 j;Vn. 8 (z) I, and if d(K 8) is the transfinite diameter of K 8, then one can easily see that g 8 (z, (0)
=
log (1 - II z)
= -
=
t!'
n2nt
= lim n.... 00 Vll1 n ' = 114
for
II z + 1hz 2 + •... H we now apply Theorem = (Pn(z)b )/(a - b), we see that D* =
1 to this function and to the polynomial p(z)
via - b II 4
for the function
.
(§ 3, Chapter VII). The convergence in this last formula is uniform in every closed subset of the domain B 8, and in particular on B. From this last it follows that ther~ exists a positive number N such that, for n > we have l;Vn. 8(Z)! > ;;;n. 8 on B. This means that, for n >N, the set E*= E(!t n .8(Z)! :5mn.8) lies entirely outside the domain B.
On (z) -
a
~
a%
~ zk • k=1
H all the zeros of the polynomials Pn(z) - ~ and Pn(z) - b lie in the continuum E, the function f'""
In the present section, we shall consider the following classes of p-valent
in such a way that R l.k
functions:
1= 1, 2,'"
Sp is the class of functions of the form
w=f(z)=zP(1 +alz+a. zi + ... ),
(1) ,
,nk)
n
~
k=1
w=F(q=~p (1 +? +?s + ... ),
"I
except for a pole at
[el.k R~. k +el.k R~. k+ ...
+ enk,k R~k.k] AlPk.
(5)
(2)
,= "".
But if we move along Gk (for k
= 1,
2, ..• , n) from w = "" to w
= 0, it
follows
from geometric coosiderationsthat th~ numbers of sheets St p lying above segments
:l; is the class of functions in :lp that do not assume the value zero in the domain
We assign to the point AZ,k (for
1
> R 2 ,k > ... > Rnk • k •
the number eZ,k> which is defined to be + 1 if the coordinate '~~ dO. p
491
Proof. from the lemma as applied to the function (2), we have, for A> 0 and
0
p> 1,
But if we set F(~) = Re itb , where ~= pe i8 , we have
o~
p
~
oR
op
(M="Rop .
211:
t\ I F (pelS) I~A dB ~ O.
(10)
Consequently, 2"
2"
0
~
SRAdlP=p ~ R),-l~~dO=t~ Lp
For \~I
\ RAdO.
> 1,
we set
[F(q]A = C"p (1
(6)
+Cl C-l+ ... ).
Then, (10) yields
00 the basis of what was said above, we then get (3). This completes the proof
co
V (n-ApL.,1l_1Dl I en I' .tt..
of the lemma.
2°. Beginning with this lemma, let us prove the following theorems: Theorem 1. lf the function (2) belongs to the class I
p
11=0
(where p :::: 1), then
1£ we let p approach 1, we obtain
co
~ n Iap +n I~ ~p+(p-l) lall~ +
11=1
... +1 ap _ll¥.
co
(7)
~ (n -
:p !
Ap) I CII I~ ~ Ap.
(11)
11=1
Proof. From (3) with A = 2, we have co
_
~O, co-I.
p~ IP-n) Ialll~
But the expansion [F (~)]), in
> 0, p> 1,
I~ I > 1
begins as follows:
[F(C)]A~C),p(1+A~+ ... +A~::::+
11=1
... ).
or co
~
(p - n) I anl~ p~ (p-n)
Consequently, c n
-1> O.
n=1
k, k + 1, ... , 2k - 1. Keeping this in mind,
=
Ap) A~
(n -
1, Theorem 1 coincides with the theorem on areas for univalent functions. Theorem 2. 1) If a function (2) belongs to the class I~ (where p ?: 1) and if
al
Aa n , where n
2k - 1:
By letting p approach 1, we get (7), which completes the proof of the theorem. For p
=
we obtain from (11) the following inequality for A < k and n = k, k + 1, ... ,
I
an 19 ~ Ap,
=
= a 2 = ••• = ak-l
that is, 9
= 0 (where k?: 1), then, for n = k, k + 1, ... , 2k - 1, we
I an 1
~ ,.----f!.. 1-
'
_\..
have
1£ we now set A = n/2p, we obtain
lanl ~ 2pn with equality holding only for a function
F(C)=C
P
(8)
(12)
Equality can hold in (12) only if c v = 0 for v -f- n, that is, only if
2p
(1 +C~r,
Ian 1~2~.
1111=1,
(9)
1) For p = 1, Theorem 2 was proved by the author [1938], where it was erroneously formulated for the class II instead of the class I{.
'1p
p
F (C) = C
(1 + C~
t.
I"II =
1.
The function (13) be longs to the class I~ because the function
(13)
._-----._._-~-~._-
492
XI. SOME SUPPLEMENTARY INFORMATION
§6. ON p-VALENT FUNCTIONS
2
~(l+C~t-=C+
By using inequalities (14) and (15) to find a bound for the right-hand side of
...
(7), we obtain immediately
Theorem 4. I[ the function (2) belongs to the class I~ is univalent in
"I
> 1. This completes the proof of the theorem. I a.g I :s;;;p (2p -
(14)
1),
,
wi th equality holding only for a function of the form
F (C) =
+tf
~P (1
P ,
([or p ~ 1), then
CX)
~ nllXp+nlll~Bp,
Theorem 3. If the [unction (2) belongs to the class I~ (where p ~ 1), then
.. ICtt I :s;;; 2p,
493
(17)
n=1
where B p is a finite quantity depending only on p. In particular, 1) p = 2,
171 I= 1.
(15)
co
~ n 1lXp+n Ii ~ 18;
Furthermore, for n = 1, 2, 3, ... ,
11=1
I I~ An,p,
(18)
(16)
(Xn
with equality holding only for a function of the form where A n • p is a [inite quantity depending only on nand p.
Proof. The first of inequalities (14) follows from Theorem 2 for k prove the remaining assertions of the theorem, we set
F(~)=CP(1 +t+~+
..·
=
r
~ (n-I)lcnlg~l.
F(q=~8(1 +t)8, 1711=1.
11=2
for all n = 2, 3, .... Since the a,., for n
easily expressed in terms of the c n, for n n
= 2,
=
Ic 11 :s 2.
=
2, 3, ..• , are
1, 2, •.• , this proves (16) for
3, "', because, in addition, the inequality yield the inequality
Jall:s 2p
as is shown by. the example of the function
p,
F(q=(~+C+ft,
and the equation
we have
which belongs to the class I aI' ••• , a
2P
-
l
~p+P (P.,-I) 4=p(2p -1),
for arbitrary c and for which the coefficients
p
can be made arbitrarily large with sufficiendy large c. Further
more, an arbitrary polynomial
Ctl ll
zP
+
a
l
zp-l
+ ... + ~ belongs to the class I
p•
H we note that the function (2) belongs to the class I~ if and only if the
function
with equality obviously holding only foe the function (15). This completes tQe proof of the theorem.
(21)
We note that constants analogous to. An,p and Bp do not always exist for the class I
In particular, since
I~ I=pCg +P (p -:; 1) c~, Ia.g I ~p I Cg I+P (P-:; 1) I
(20)
11=1
with equality holding only for a function o[ the [orm
CX)
c1 = a/ p
(19)
oX>
~ n!lXp+n Ii ~ 300;
1'1
Ie n I :s 1
171 I = 1;
2) for p = 3,
for > 1. Then, from the lemma, with A= 2/p, we obtairi, by letting p approach 1 from above,
It follows that
+tr,
F (~) = ~i (1
1. To
n
1
F(
=zP(l +at z + ... )
-- __.
··
••.
_r_"K'.~"'"""'""""'-"""'"_=,,_._·-
.._..
_.~_._.'_.~.
_
~._.:..;;~.:;;;:.:,:~~;;.
..~:.-.._.~.
':':'::~~''':;-:::,",~~_;:
•
--
494
~
• .::
;-'-'~'
.;:;;:;::--.,.
'.T
_
.~,~
'"'
. _ '__
:.'_~:'-:_.'
-
~..
-
',:-"~:",
-.
__ • __
~:.: "''""'.~ ',r
.:'.;""....
XI. SOME SUPPLEMENTARY INFORMATI0N
§6. ON p-VALENT FUNCTIONS
belongs to the class Sp and if. we note that, for a 1 = a 2 = ... = ak-l = 0 (k:;:. 1), we have a 1 = a 2 = ... = ak-l = 0 and a" = - an> where n = k, k + 1, ... , 2k - 1, then, in accordance with Theorem 2, we obtain the corresponding theorem for the
cf(z) _
c - f (z) -
z
P+ (
ap
+ 1-)c z i P+
495
...
also belongs to Sp. Therefore, by Theorem 5,
class Sp: Theorem 5. If the function (1) belongs to the class Sp ([or p a 1 = a2 = ... = ak-l = 0 (for k? 1), then, with n
?
lap++1~2, lapl~2,
1) and if
k, k + 1,"', 2k - 1, we have
=
,
lanl~2pn'
so that
(22)
1+1~4,
Equality holds in the above relationship only for a function of the form
zP
F(z)= (1
+
that is,
~' 1"l1=1,
(23)
1
Icl~4'
1JZn) n
belonging to the class Sp' In particular, for an arbitrary function (1) in the clas.s
30. From what we have already shown, we can obtain some results concerning coverings.
belonging to the class Sp (where p
? 1)
lR completely covers the disk
maps the disk
IwI ~
1/2 P
Izj < 1.
f(z) =
l1,
=
In the case of the func
in addition to the preceding, a p = 0, then, from (26), we have
lei?
n,
1/2; that is,
Now, let the function (1) be any function in the class Sp. Then, the function
+ 1;
though not always any larger disk with center at w
Iwl < 1/4.
tion (24) the surface !R does not cover the point w = - 1/4. This proves 2).
surface
Then,
= '" = a p- 1 = 0, then R completely covers the disk
(26)
lR completely covers the disk \wl < 1/2. In the case of the function (25), the R does not cover the point w = - 1/2. This proves 3).
Theorem 6. Let !R denote a Riemann surface onto which a function (1)
2) if a 1 .= a 2
!R completely covers the disk
This shows that
Sp, we have la 1 1 ~ 2p.
1)
.,'~ .•~._ .,..?-__, _,"_;:;;""' ~ . -;;"'.,.•;'" '_'"''"'''.'-''''''__'''
.............
f 1(z) =
\wl < 1/4
0 since the function
zP _mi=ZP+2z iP + ...
p+1
V I (zP+1) =
zP
+ p+ I z'P+1 + .,. a
(27)
also belongs to the class Sp. This is true because, if there were points z l' (24)
z2' ••. , zp+ 1 in the disk
Izi < 1
at which fl(z 1) = fl(z2) = .•. = fl(zp"t1), we
would have belongs to the class in question; 3) if al
=a2 =
••• = a p
= 0,
then !R completely covers the disk
though not always a larger disk with center at w
=
I(z\p+1)= I
Iwl < 1/2
p+1 (Z2 )=
... =
I
(zpp+i + 1),
0 since the function
so, that by virtue of the p-valence of the function f(z), among the numbers
f(Z)=l ZPZiP=ZP+Z3P+ '" belongs to the class in question.
(25)
z~ + 1, zi + 1, z~ + 1
••. ,
= zi + 1.
z~ ~~, there would be some that are equal. Suppose that
Then, z2
only terms with powers
= Tfz l '
zn(p
rf+l = 1. But since the series (27) contains
+ I)+P, for n = 1, 2, .,. , we would have
Proof. We begin with part 2) of the theorem. Suppose that the function
11 (z~)
I(Z)=ZP+apZi P+ ... belongs to the class Sp. Let c denote a value that f(z) does not assume in
Iz I < 1.
Then the function
and consequently the result Tf
= 1;
rf = 1.
= 11 (1j Z1) = "lP/1 (Z1)
From the equations
that is, z 1
=
if = 1
and TJP + 1
z 2' which proves the p-valence.
= 1,
we obtain
~--- ...... ---
496
§7. ON TIlE CARATHEODORY-FEJER PROBLEM
XI. SOME SUPPLEMENTARY INFORMATION
Now, if c is a value that f(z) does not assume in the disk
Izi < 1,
then
which belongs to the class
!;
497
for arbitrary·. p> 1.
f1(z) does not assume the valueP+..jc. Consequently, in accordance with 3) we have
P+~ ~ 1/2, that is,
Ie I ~ I/ p P + 1.
§7. Some remarks on the Caratheodory-Fejer problem
This completes the proof of the theorem.
and on an analogous problem 1)
4°. Let us find another bound for the absolute value of the derivative of func tions belonging to the classes
fu subsection 1° of the present section, we shall give a simple derivation of
!p and !;.
Theorem 7. If the function (2) belongs to the class in the disk
1(1
!p (where p
~ 1), then,
> 1, we have
the results of Caratheodory and Fejer 2 ) dealing with the problem of the extend ability of a polynomial in z (by adding terms of higher powers of z) to a power series representing a rational fraction with constant modulus on the circle
I p' (1:) I :s;;;;; c (lXI,
Clg, ... , IX p _ l )
Iz I ==
1
and on the minimal property of the maximum of the modulus of such a fraction in
ICI~IP I • IC
the· open disk
Iz I < 1
among all functions that are regular in that disk. This
derivation is based on simple results dealing with the solution of the problem of coefficients in the case of bounded functions. In subsection 2°, we consider the
where c(a 1 , a 2 , ••• , CIp-1) is a finite quantity depending only on al' a 2 , ••• , a p - 1' If the function (2) belongs the class (where p ~ 1), then, in the ais.k
extendability of a polynomial in a similar fashion, starting with results dealing
1(1) 1,
with the problem of the coefficients in the case of functions with bounded mean
!;
values of the modulus on concentric circles. Here, the minimization of the maxi
I CIP
I F' (C) I ~ c (P) I C1_ 1 '
mum value of the modulus is accordingly replaced with minimization of the maxi mum of the mean value of the modulus.. A particular case of the extendability of
where c(p) is a finite quantity depending only on p. Proof. From (2), we have, iii the disk
a polynomial that is considered in subsection 2° has more than once found appli
1(1 > 1,
cation (by Landau, Fejer, Szasz, and others) in inequalities relating to bounded 00
IP
(C) I ~ pIC Ip-I + (p - 1) !IXIIIC IP-II +
... +
1 1Xp-1 I
+
!
n 'I ~I~':!
11=1 00
=pIC\P-I+(P-l)l lXlIICI P-II + ... +llXp_d+] l~l·vnIIXJI+/l1 ~p ICI P- +(p -1) 11X111 CID-\I+
1
... + IlXp _ll 00
11'
!
n IlXp +II Ill,
11=1
We note that the order of the inequalities given by Theorem 7 is sharp since
F(C)=P
CI-l
PC
_ 1'
obtaining sharp inequalities in all cases, thus removing this gap at least theoreti
of the solution given by I. Schur to the problem of coefficients for bounded func
from which Theorem 7 follows by virtue of Theorems 1 and 3.
for the function
section 2°, our second method of extension enables us to indicate a path for
tion of the Caratheodory-Fejer problem, give a simple derivation of the final form 00
1l=1
pi F'(p) = pl-l
the particular extendability referred to does not exist. As will be shown in sub
cally. We note further that in subsection 1° we shall, by beginning with the solu
11=1
+ 1/ 1: I C12~1I
functions. However, the problem of sharp inequalities has remained unsolved if
tions in the case of interior points of the domain of the coefficients. We shall use the following notation: 3 ) We denote by B the class of func
+ C lZ + •.. that are regular in the open disk Izi < 1 and that satisfy in thaC disk the condition If(z)l:s: 1. We denote by HI the class of func tions f(z) == Co + C 1 z + ..• that are regular in the disk Iz I < 1 and that satisfy the condition tions f(z) ==
Co
1) Goluzin [1946a].
2) Caratheodory and Fejer [1911).
3) O. Beginning with the con Consider the space
tR n ,
cept of a neighborhood, we define cluster points, interior and boundary points of a
to (2),
set, open and closed sets, and convex sets in the usual way. The space ~n can,
+
f. (.{) = E, f k :...! (z)
in case of necessity, be regarded as a 2n-dimensional Euclidean space with
Ik-l (0) zlk (z) 1+Ik-l (O)z/k (z) ,
Cartesian coordinates corresponding to the real and imaginary parts of the nwmbers
Co' Cl'
••• ,
Cn- l'
We denote by B (n) the set of points (c 0'
C
l' ••• ,
C
n-I)
€
tR n
such that the
we can show by induction that
numbers co' cl' " ' , c n- I are the first n coefficients of some function in the class B. We denote by H I(n) the analogous set for H l' These are bounded sets in ~n because, for all k
=
0, 1, ... , we have
Ie k I ~ 1
for both the functions in
B and the functions in HI' by virtue of the integral representation of the C k. Since the principle of compactness holds for functions in Band HI and since the limit. of a sequence of functions be longi ng to either of these c lasses also belongs to the same class, it follows that B (n) and H ~n) are closed sets. Further more, since for functions fI(z) and f 2 (z) belonging to one of these classes the functions AfI(z) + (1 - A)f 2(z) for 0 < A < 1 also belong to the same class, it follows that B (n) and H ~n) are convex sets. It is also obvious that the origin (0, 0, ••• , 0) is an interior point of these sets since the polynomials Co
( )
J .-k Z
nator by 1 +
Z
ClO
+~n sz + ... + ;;ozn-l + 1Z +... + n_1Zn- 1 . Cl
Cl
Proof. 1) Let us suppose that a function fez) €
B other than a fraction of the
1) The basic idea of the proof is borrowed from Schur. See Schur [1917] and also Bieberbach [1927].
'a
( 1)
1, k
= 1,
2, ... , .
(3k. In particular, fo(z)
=
f(z) is also
+ ... + zn-II-1 and this, by the hypothesis of the theorem, is
impossible. Since
Ifk(o)! < 1
for k
=
0, 1,'" , n - 1, the functions (2) are defined
for k = 0, 1, ... , n - 1 (that is they have denominators not identically equal to
0). For these functions, the inverse formulas
Ik dO)+zlk(Z) k=n-I, ... ,1,Jo(z)=J(z) I Ik-l (0) Zlk (z) ,
+
(4)
are also meaningful. Now keeping the values h(o) for k = 0, 1, ... , n - 1, let
us calculate the functions
Cl n 1
I Ek I=
equal to 1 because otherwise we could arrange for
this by changing the arguments of {30"'"
C1 z + ... + Cn- 1zn-I belong, for small Cko both to B and to HI' B(n)
£k
,
a fraction that reduces to the form (1) when we multiply its numerator and denomi
Jk-l (z)
there correspond in B only fractions of the form
~k+~k_1Z+'''+~ozk
= Ek ~o + ~lZ +... + ~kzk
Here we can assume the
=
10 • Theorem 1 (Schur). To points (co' cl"", Cn-I) on the boundary of
(3)
f :(z),
for k
J:-I (Z)=Jn_I(O), J't-I(Z)
I All the
f :(z)
=
n - 1, ... , 0, from the formulas
Ik_I(O)+zl:(z) k=n-1, ... , 1. (5) 1+ I k - 1 (O)zl% (z) J
belong to B. From (4) and (5), we have
ik-I(z) - J:-I (z)
(1 -Ilk_I (0) II) z (fk(z) (1
n (z»
+ I k - l (0) zlk(z»(1 +I k_1 (O)zll (z»
Ilni\/pr~av
nf
Z (Ik (z) -
f%(z»
rpk
W~~hinO'tnn
(z),
I ihrarv
--~
500
§7. ON THE CARATHEODORY-FEJER PROBlEM
XI. SOME SUPPLEMENTARY INFORMATION
where the ¢k (z) are, for k = n - 1, ... , 1, regular in
I(z) - It (z)
1z I
to some positive q is a function f(z) € B, then there exists a q '< q to which
< 1. Therefore
= zn-l 0,
1 ,
=
0 are equal to
co' .•. ,c n - l ' These fractions include the fraction R(z, co' .. , , c n -
1 ).
If we
cross-multiply in the equation
- + +-aozn-l _ n a o + ... + n z
an 1
'"
Cl _l
1 -
= -
Yk =
f3k
+
(n - 1), "', - 0, 0, ... , n - 1 and Ok = i ({:3k - f3.-k), for k =
0. 0, ., . , (n - 1) is also a system. Here, at least one of these
=
Y-k and Ok
=
8-k>
this proves the assertion made. Thus, we have
shown that corresponding to every positive root of equation (7) is a fraction (9)
the first n coefficients of which in the expansion about z
A
also provide a system
solutions is other than the zero solution because, if this were not the case, the ditions Yk
aoz n I 1
130' 13-0' ...• 13-(n-1)
system (:3k would also be the zero solution. Since Yk and Ok satisfy the con
Proof. We seek all fractions of the form
A an_I
for k
- (n - 1), .. , ,-
1=0. Cn _l
(3n-I"'"
of solutions, as one can easily verify. Therefore, each of the systems
of the absolute values of the real roots of the equation 1) of degree n
-
tions, then the numbers
Co
+ + CI Z
with the property required. Suppose now that all distinct positive roots of equation (7) are constituted by Ao"'"
Av , where Ao > ... > Ay and suppose that corresponding to them are
the fractions Ro(z), ••. , Ry(z) of the form (9). Let us show that of these\only
Ro(z) is regular at •••
and equate coefficients of like powers of z we obtain a system of equations for
the roors of equations (7) and (8) are real, bur this factor is of no significance ro us.
We note first of all that the fractions Rk(z), for
k = 0, 1,"', v, have no zeros or poles on the circle Izi = 1 since \Rk(z)! = Ak, for k = 0, 1,···. v, on Izi = 1. Suppose now that Rko(z)'is regular-in the . closed disk
I) FuUhermore, by virtue of the familiar properties of characteristic equadons, all
Izi < 1.
Iz I 5: 1.
Let
R k 1(z) denote an arbitrary fraction among the fractions
R k (z). The difference if; (z) = R k 0 (z) - R k 1 (z) has an expansion about z = 0 that begins with terms of power no less than n. Therefore, the number r of roots of
_ _ _ _ ilfima
'lIIIl~
_!IIi
_:wwm
1!!_~~~_'la\Ilo~_~milll!i;;l1J1l1U'H_~f~~·~ikl:I~;·~II~,y~~~i'I"-'QIo~~";i:.y.~~.;I_~!iiZ_II!l_~
_ _~ ~ R I I : I < 1 l i ~
----"
504
S7.
XI. SOME SUPPLEMENTARY INFORMATION
¢ (z) (counting each root according to its multiplicity) in the open disk
Iz I < 1,
and the number a of poles in that disk does not exceed n - 1; specifically, it is equal to the number of all the poles of Rk 1(z). Furthermore, on
I~ (z) I~ II Rll o (z) I-I Rill (z) II = IAllo Consequently, ¢(z)
Izj =
t.
0 on
Izi =
1. Therefore, as
Iz I =
1,
Alii I -=F 0.
z moves
arg R k /z) acquires an increment of 217(t - 0)' where t is the number of zeros of
jz I < 1.
This number does not exceed n - L Therefore, the increment
in arg (¢(z)/R k1 (Z» as z moves around Izi = 1 is equal to 217(r- t) ~ 217. Con sequently, the circle Izi = 1 includes a point Zo at which ¢(z)/R k l(z) is posi
final form of the solution by Schur of the problem of coefficients for bounded func The~em 4 (Schur). For the point (cO"'" cn-I) to be an interior point of B(n), it is necessary and sufficient that the inequality
O
I Rll o (zo) I =
I rp
(Zo)
+ Rill (zo)
1=
1 Rkl (zo) I
I1 + R:~(~o) I > IRkl (zo) I= Akl'
°1
° Dk(l)= I ° ° ° ~o
tive. We have
AII =
505
tions in the case of interior points of the domain of the coefficients.
1 around the circle
1, the function arg ¢(z) acquires an increment 217(r- 0), and the function
Rk 1(z) in
ON THE CARATHEODORY-FEJER PROBLEM
~1
~o
~k-l
~k-'j
.
Co
"
...
Cl ••• CII_l
° .Co ... ... 1 Co 1>0, k=l, ... ,n, ° °° ...... °
... 1 ° ... 1 •.• ° . . . °. ° ... ° ° ... 1 CII_i
(11)
~o
be sa tisfied. Proof. We note that the determinant Dn (A) (d. (7» is an even function of A because, if we divide each of the first n rows and the first n columns by - 1, we obtain Dn(A). Therefore, the determinant D n(I) is of the form (11).
> Ak l' This shows that Ak 0 must be equal to AO; that is, the fraction (9), which is regular in \zl < 1 corresponds to the greatest root of equation (7).
(c 0' ••• , c n-I) lies inside B(n), then the point (qc 0' ••• , qc n-I) lies in B(n)
Therefore, R(z, co'"', c n-
for some q
that is, Ak 0
1)
= Ro(z) and A(co"'"
c n-l) = AO' which proves
> 1. Consequently, there exists a function f(z) € B whose expansion
begins qc 0 + .•. + qc n_ I z n-
the fir st part of the theorem. Now, in the case of real co' " ' , cn-I' if we set ak = Xk + iYk, the system (10) can be broken, by separating the real and imaginary parts, into distinct sys tems each consisting of n homogeneous linear equations in the unknowns Xk for one system and Yk for the other. Here, the determinant of the first system is the determinant (8) and that of the second system is d(-A).
We first prove the necessity of the conditions of the theorem. H the point
For the system (7) to
have nonzero solutions, it is obviously necessary and sufficient that one of these smaller systems have nonzero solutions, that is, that A be a root of the equation
d n (A) d n (-A) = O. Furthermore, just as in the proof of the first part of the theorem, we can show that A must be the greatest root of this equation, that is, the great
1
+ .•.. On the other hand, the function f 1(z) = I f(z)/q = Co + ... + c n- I zn- + ... satisfies in Izi < 1 the inequality Ifl(z)1 ~ 1/ q. It then follows from Theorem 2 that A(c 0' .•• , c n-I) ~ 1/q. But then, in accordance with Theorem 3, Dn (A) has no roots in 1 ~ A < + 00. Since Dn(+oo)=+oo,wehave Dn (I»O. Furthermore, since (c o,···,c n-l)isan interior point of B(n), it follows that the point (co',··, Ck-l) is obviously, for k < n, an interior point of B( k) in the space 'R k • Consequently, by analogous reasoning, Dk (l) > 0 for k = 1, 2, •.• , n - 1. This proves the necessity of con ditions ( 1 1 ) . ' ~ We now prove the sufficiency of the conditions of the theorem. H conditions,
est of the absolute values of the roots of equation (8). This completes the proof
(11) are satisfied, then the equations Dk (A) = 0, for k = 1, •..
of the theorem.
roots in the interval 1
Before turning to the following problem of the Caratheodory-Fejer type, we shall give an application of what we have just discussed to the derivation of the
~
A') ple roots in 1 S A < +
00.
Furthermore, the interval 1 S >..
.. **. Since by assumption this is impossi
(Here
f(O
ble, it follows that the equation D jJ-(>") = 0 has no more than one root in the inter
k=O
has no zeros at all in I S A < +
00,
> O. Consequently D jJ-(>") = 0
which contradicts the assumption. This con
tradiction shows that Dn (>..) = 0 has no roots in the interval 1 S A < + fore
A(C o"'"
cn-I)
00.
Theorem 1, the point
(c 0' ••• , c n-I) is an interior point of
B(n).
This com
pletes the proof of the sufficiency.
k=O, 1, ...
(2)
_1 \ f (C) ( 21ti J V Tn-l
+
..•
+ T rn-t) dC.
(3)
0-'
But, in addition to formulas (2), we also have the formulas 1 21ti
Cf
.)
k
(C) C dC = 0,
k=O, 1, '"
I C1=1
is an example of a function in the class B that is different from a fraction of the form (1) and that has an expansion beginning z 0 + ..• + C n-I zn-I + .. '. By
dl.,
ICI=I
There
< 1 and, consequently, the fraction R(z, co'"', cn-I)
Ck+1
means the limiting values mentioned above.) On the basis of this,
vallS A < +
and D jJ-(I)
J
I C1=1
n-I
~ Tk Ck -....
00
\ fCc)
~ 21ti
DjJ-_I (A **). But D ~ (>.. *) and D ~ (>... *) obviously have the same sign. Therefore
However, D jJ-(+ 00) = +
and it can be expressed in terms of these
_ -
DjJ-_I(>"*) have the same sign, as is true, analogously, for DjJ-(>"**) and D jJ-_I (A *) and' DjJ-_I(>.. **) must have the same sign. This leads to the fact that
1 along all non
limiting values in accordance with Cauchy's formula. Consequently, we obtain
for the coefficients c k the integral representations
consecutive roots and use (12) and (13), we can show that D~ (>.. *) and
00.
Iz I =
has definite limiting values almost everywhere on the circle
This enables us to replace the parenthesized expression in (3) with an arbitra~y, function, regular in the closed disk first n coefficients equal to
"I
Yn-l' •••
S 1, whose expansion about
,=
0 has its
'Yo' For such a function, we may take,
2°. Theorem 5. Corresponding to the points 0, possessing the property (the support plane) that
=
in the portion of space defined by the inequalityI~~Io(akxk + (3kYk)
••• ,
Cn-I)' Consequently, this
Jj
HI
r-.l 0
0
< 1.
We
f (z)
= 2~ ~ \f (Q d~ I= 2~ ~ IP (Q + ~n Q ~) II d~ 1 ICI=1
is
ICI=I
lei - -2lt j IP(qIIIP(QI2_~nQ(QP(~HldU ICI =1
extremal and hence is of the form (7). This completes the proof of the theorem.
~ 2~ ~
Theorem 6. For any polynomial Co + ciz + ... + cn_Izn-II= 0 there exists a
unique polynomial of the form
ICI=l
+. "+ ~n_Y-IZn-H) (~n-.-l +...+~ozn-'-I), v-n-l
""""
8
_
1
-21t
Izl
-t
2{
1":n~oInCnpn =
Since the i!1tegrand is nonnegative, we have
Yo
< P < r < 1, we have formula (8), which we
0), we can prove the formula
---_~~
pn= ~ C n n 27tt"
00
m
=
f(~) [2m (_I ~ nc-n)_ ~ _ 10 .i.J InP
~ I c
hold in the disk
Proof. Again assuming that 0
Proof. Let us suppose that condition (7) is satisfied and that fCz) is a
(12)
n=O
v~1 00
:s 1.
\yAl)!
and
complex n umbers such that For the inequality
Yh
1
lyA 2 )j )
1= 0,
(n = 0, 1, ... ) denote two sequences of
Yh2 ) = 0,
and
L;::'=01 yP)1
+
L;::'=olyF)1
1 -1
'Y,
0
n=O
Incn
E a,
(11)
I.~o T~"·I + I.~o TA",·I.;; IT~"l to hold for an arbitrary function fez) = L;::'=oCnZ n E Be, it is necessary and
(141
sa.
XI. SO\lE SUPPLEMENTARY INFORMATION
520
SOME INEQUALITIES FOR BOUNDED FUNCTIONS
521
that is, to the inequality
sufficient that ~-
ro
ffi
i l'n~n-m ~ i
ro
(-~ ~ "l'(ll~n) - J..:::> )'(1) "'" In 2 0--
o n=O
I_I ~ "l'(~)~n I )'(1\ " "
n=O
0
In
I ~ I=
on
1.
I
(15)
n m InC -
I'
n=m
I
~
I 1= 1.
on
n=m+l
(18)
The same is true for condition (12). By an analogous transformation, we can show Equality holds in (14) only for functions f(z)
= cO' where \c 0 I = 1.
that condition (15) is equivalent to the condition
Proof. Inequality (14) is satisfied if and only if the inequality
In~o l'~)cn +
e
ia
n~/~~)Cn 1= r n~o(l'~) + eial'~) Cn r ~ 1l'~ll I
V
_I )'(1)
n=O
o
(16)
is satisfied for all real a. But, for fixed a, this last inequality is satisfied for
2 1/
all functions fez) € B if and only if (Theorem 1)
"'"
"l'((lcnli In
(1(11)
l: (l'~l + ei"l'~»Cn) ~ ~
on
o n=O
I ~ I=
Consequently, for inequality (16) to hold for all fez) E B, it is necf'ssary and
n=l
In
)'(t)
0--
0
"l'(llen
~ In n=!
I~ on
le1= 1.
(19)
2°. As an application of Theorems 1-4, we present Theorems 5-10. Theorem 5. 1) For all functions fez) = I;:'=ocnz n E B in the disk Izl..::: (m + 1) _ 1 but not always in a larger disk, we have
l
j t<m (z)
(17)
l.
1
)'(t) "'" 0
ro
ffi
_21_ ~ "l'(2)cnl~I_~ ~
+ n=! f (n + 1) (n + 2) ... (n + m) cm_nzn I~ mI.
(20)
= c m zm, where Ic m I = 1. f(z)= I;:'=mcnzn € B in the disk Izl..::: 2 t1 (m+1)_ I
with equality holding onl y for f (z)
sufficient that ine quality (17) hold for all real a and this is equivalent to the
2) For all
fun~tions
but not always in a larger disk, we have
condition
ffi
If(m) (z) I ~m!,
(J..~ "l'(I)~n) _I-~ V )'(1) "'" In )'(1) "'" In
"l'(21 Cn
o n=O
0
n=O
I
~2
on
I ~I= 1.
(21)
with the same extremal function. Proof. To prove 1), we need to investigate whether the condition of Theorem 1 holds. with
which is what we needed to show. In accordance with Theorem I. equality holds in (14) only when f(z)= co' where
\col
l'n=n(n-I) ... (n-m+I)zn-m
1. This completes the proof of the
=
theorem.
(n=m,
m+I •...).
In this cas e, condition (4) or condition (18), which is better. takes the form
With an eye to applying these general theorems to particular cases. we note that it is convenient to write condition (4) (condition (2) in case m
=
0) in a
1.-
different, equivalent form. Specifically. condition (4) is equivalent to the inequality
m dm
mJ dz
m ~ znCn- m I IJ..- d m ~ "'" :: ml dz "'" :::>,
n=m
n~m
znCn- m -
1
I
on
ICI=1.
But
~"l' cn-mi~
_I )'m "'" In
I
n=m
~
1_
I
co
~
-.y;; "'" rn~n-m n=m
I~
on
ICI=I,
m
co
co
I d ~ nCn-m _ I d m ~ ",I dz m £.J Z -",I dz m ~
n=m
n=O
nrn
z..
m
=
I d m C··m I "'! dzm 1- zf= -,,--
.,_~..
(22)
522
§s.
XI. SOME 3YPPLEMENTARY INFORMATION
Therefore, inequality (22) takes the form
1(I =
1 CI =
(23)
1.
1, we have
1 -I (1 - Cz)m+ 1 -
1 I~ 1 - «1
+ Iz )mH -
1) = 2 - (1
+ Iz I)m+1,
and this inequality is sharp (equality holds only when (= - e-ia,gz). Con sequently, inequality (23) is satisfied if and only if 2 - (1 + Izl)m +1 20, that is, if and only if Iz I :s 21/ (m +1) - 1. By Theorem 1, this proves 1) of Theorem 5, m including the uniqueness of the extremal function f(z) = cmz , where Icml = l. To prove 2), to the class
I)
we need only note that the function f(z) = l;::'=mcnzn belongs
closed disk Izl:s rm· Equality holds only for f(z) = co' where m, the number r m cannot be replaced with a larger number. 2) For all functions f(z) E
Proof. Let us apply Theorems 1-3 with Yn Yn = 0 for n
> m. Inequality
I~ml
l: (n+ 1) .. . (n+m)z1IC -11
zm- 1Cm - 1 1 on
ICI=
1.
(24)
1
where \(1 = 1, and its sharpness for even m (equality holds when (= - e-ia,gz)
II
imply that inequality (24) holds for all z for which 1 - Iz I
But this condition is identical to inequality (22) and is satisfied only for
But r m
< r m +1
(for m
2 Iz 1m 2 0, that is,
2, 3, ... ) since
=
1, which proves part 2). This completes the proof of the
2r:+1 +rm+I-1
theorem. In what follows, we denote by sm(z) and rm(z) the mth partial sum (begin n ing with m = 0) and the corresponding remainder of the series I;::'=ocnz : l;::'=mcnzn.
H lim m--> 00 r m
=
> 2r:.:/=1 + 2rm+I-1 = 0 and 2r: +rm-1 =0.
q < 1, then, for all m = 1, 2, ... , we have r m < q and, con
sequently, 2qm + q - 1
> O. By letting m approach
00
in this last inequality, we
see that q 2 1. Thus q = 1. This completes the proof of the theorem.
Theorem 6. Let m denote an integer greater than 1. Let r m' with 0 1, denot e the (unique) positive root of the equation 2r
-
for Izl:s rm. Here, when m is even, we cannot replace rm with a larger number. I) By virtue of Theorems 1-3, this proves all the assertions of the present theorem.
on ICI= 1.
=
on Icl=l,
1
11=0
l:~~cnzn, rm(z)
(18): which with m = 0 is equivalent to inequality
- zm- 1Cm- 1 m m m ~ 1 - I z 1 - I z I(l + I z 1 - I ) = 1 -I z 1- 21 z ,
OJ
=
and
The inequality
11=0
sm(z)
= 0, 1, ... , m - 1,
11 +zC+ ... +zm-lCm-ll~lzC+... +zm-lCm-ll
[I - cmz mI-I z 111
(n+ l) ... (n+m)zIlC Il I
Iz 1:s 21/ (m + 1) -
forn
1.
(2), is of the form
11 - cmzml ~ I z 111 -
OJ
~ I ~!
= zn
=
that is,
in the cl ass. B. By Theorem 1, this requires that we investigate the condition
l:
1. For even
the inequality ~Hsm(z)) 2 0 holds in the disk
3) For f(z) E Be, we have sm(z) E G in Izl:s rm. We note that r2 < r 3 < r 4 t
2) If t 1 and t 2 are two points of discontinuity of the function a(t) such that a ::; t 1
C eit+z
z/' (z) _ j(z) -
( .(z), like ( (z), is a function in the class Eg' Formula (7) is the first variational
in
then, for suffic ien tly small real A, the function
(1* (t)
y"p (t'A),
e;: t
(16)
,
author [1938] that the inequality
An analogous situation occurs in the problem of maximizing
Pi (t) = t (1 -
T
Proof. For the functions F«() mentioned in the theorem, it was shown by the
1 that corresponds to boundary points on
Pi (t) =
e i '"
where a is a real number.
+rtF'
< Al < 2,
P(t)=t+
537
I and where a
I(I
close to I,
< Al < I in the case of this extremal function. However, for
I(,
suf
ficiently large, one can show by a roundabout procedure that the function F«() = ( + e i a(-1 is an extremal function. This fact is included in the following distor tion theorem. Theorem 5. For a function F«()=(+atl~+a/(2+ ... EI' the sharp inequalities
1 I~
W
in
It I> 112 + 1.
This inequality leads to inequalities (14) and (IS). This completes the proof of the theorem. It is still an open question whether inequality (IS) is valid in the entire
I(I
> I under the conditions of Theorem 5. The corresponding question with regard to a bound for [arg f'(z) I in the class S· has an affirmative answer, domain
~" Th is wi 11 be shown below,
4°. Let us establish a theorem for the inverse of a starlike function. Although, as we have mentioned, the application of the variational method set forth here
hold for form
1(' >.J2 +
to the particular question of solving this problem is not new, we shall nonethe
1 + W'
(14)
• 1arg F' (t) I ~ arcsin W1
(IS)
IF' (t) I,,;;;;; 1
less give its solution, treating it as an application of the general Theorem 2 of subsection 3°. Theorem. For given z in the disk
Izi < 1,
the function arg f'(z) in the class
S· attains its maximum and minimum only in the case of functions of the f(lrm 1 with equality holding only in the case of a function of the
---l)~e
the author's article [1945] and also Theorem 5 of
§a,
Chapter Xl,
XI. SOME SUPPLEMENTARY INFORMATION
538
§9. A METHOD OF VARIATIONS
z f(z)=-,-, -_10_'. ,
corresponding to the numbers (1)
4
> O.
The problem
ffi
that we are considering is a special case of the problem of Theorem 2 of sub
t
(! Ck ) = !
COS
< 11 no more than two points of discontinuity, which must be roots of the
ffi (C) = ffi
equation e-i/ r
F (t)= ffi ( 1 ~ e
If
r
+C
e- it r (I _
e
it
)_
r)2
-
(this is equation (10) of subsec tion 3° with IlJ (w)
=
J..!!L
_
0, C-
± iw).
rj'(r)
(2)
rn (llog (1 --e- II. r) -
of the intervals (t I , t 2 ) and (t 2, we reduce it to the form
H there are two poin ts
ire-it
C 1- e it r
)
t
,=
e
,
~ ffi (i log (1
at
2r a)
rC
-
at (llog (1
+ Ce-
It
(1 - eit r)i) = 0
+ 2r + r a+ 2Cr - elf (r' + Cri» = O.
III
= O.
This equation must be satisfied by the numbers 'k = e it k , for k = 1, 2, 3, 4,
ffi (
e- it r
1 - e If r
)
=
r-cost
11 - e it r
-
e-itl r))
> at (llog (1 -
e- ita
12 '
r».
"
z
O~A~I,
we obtain
---l..!:Il.- =
C- r/'(r)
(4)
1~
Itj 12 1 "' r A l+e "' r +(l-A)i_e- r it2 l-e I l r
Therefore, by setting
we write this equation in the form of a fourth-degree
(1 + 3r' + C+ Crg) c + agC' + llaC +
e- it r»= _
< t < t 2• But then, since
(l-e "lz)2A(1_e-it2z)i(1-A) ,
Ck -
for t l
Furthermore, remembering that the extremal function r(z) is of the form
f(z) =
9
>r
we conclude that the first term in (3) decreases in the interval (tl' t 2 ), so that
algebraic equation I
r.
> r, k :-1, 2, 3, 4.
It follows, in particular, that cos t
=
ffi (e iU r - elf (1 + 3r 9 + C+ Cr'1) + 2r + r a+ 2Cr) = O. Then, by setting
+
COS k
H we now replace the first two terms with their 'complex conjugates, which does not change the real part, we obtain the equation
it
(JJ!L)::>: 1- r rj'(r) :O----1+r"
k=l
and we can then reduce this to the form
r - e- (1 + C+
we have
2 t ::>:~+3 r +1+r -3+ r+, (l-r)2(2+r»3+ ..:;.. cos k:o---- r r 11-r +r r (l r)
(3)
l/J'(t) has one zero inside each t l + 211). By ridding equation (2) of denominators,
ffi (e- lf (1 - e- it r) (1 - elt r)'
ffi (e-
Izi < 1,
This shows that
< t 2• Then, the function F(t)
it
+ 1 ~r2 ffi (C).
4
Let us suppose that there are two such points of discontinuity t I and t 2'
iit
3r
"'\"1
(d. (10') of subsection 3° wi th llJ(w) = iw) has the same value at the two points.
indexed so that t l
=++
Consequently,
of discontinuity, the function
~ (t) =
tk
k=1
On the other hand, since ~(f(z)jzf'(z) > 0 in
function a(t), correspo"nding to an extremal function r(z), has in the interval 11 .:::;
4
k=l
section 3° when IlJ (w) = ± iw. It follows from the proof of that theorem that the
-
mentioned above.
tk
We now have from (5)
where a is a real number.
Proof. Without loss of generality, we may assume that z = r
539
(5)
we have
1 +e-itkr l-e -itk r '
k= 1,2,
(6)
540
§9. A METHOD OF VARIATIONS
XI. SOME SUPPLEMENTARY INFORMATION
(C 1 ie~:ll~l r)- m(C_l_ie_~_it"II~:-1r-) -rn(tc( e- r _ e-1t'llr ))-l-rn(t l-e-1t1r l-e-It'llr - 2 -
where a«() is a real nondecreasing function in the interval 0 S () S
ffi
a(17) - a(o)
I C IS - = -23(AC.C~ _I C 1'11 3 (1 +e+lt!r
-
2
1
2
S
(z, t) da. (t),
(3)
where
_I C I'l 2r (l-r)(sint1 -sin t a)
l-e+it!r 'l-e-itSr -
~ -1
(1 - A) CiCg) =-2- 3 (C.C~)
its 1+e- r)
Il-i/1rll'll-e It'llrll
s (z, t) =
(f' (z»)
and
0, which is a 2nd-degree algebraic equation:
ffi ( '(f(z)) (1 -
2. For a given entire function ¢ (w) and a given point z in the
O. Consequently, the only
possible points of discontinuity of the function a(t) are the endpoints the roots of the equation F(t)
=
If(z) I is equivalent to the problem of maximizing I (f(z))\ with
Theorem
then a(t) = const between any
two roots of the equation F(t) = ~('(f(z))«z, t))
543
+ Z2)~) =
in the class
0,
Tr
IIf> (/' (z» I
is attained only by functions of the form
(9)
so that the number of these discontinuities does not exceed 4. H there are more
and
3
f(z) =
~ AnS(Z, tn)' n=1
3
An~O,
~ An = 1, n=1
than two points of discontinuity, denote three of them by t 1, tz, and t 3 , indexed in such a way that - 1 $ t 1 < t 2 < t 3 $ 1 and F(t 2) = O. By applying formula (6),
where t n belongs to the interval - 1 $ t $ 1. If'e exclude from consideration the
let us show that R('({(Z))S(z, t)) has, as a function of t, equal values at the
case in which '{f'(z)) = O.
points t l' t 2> and t 3' But then its derivative has a root inside each of the inter vals (t1' t 2) and (t 2, t 3 ); that is, equation (9) has three distinct roots. This con tradiction shows that a(t) does not have more than two points of discontinuity in
f (z)
the interval - 1 $ t $ 1, and this means that the function
is of the form (8).
This completes the proof of the theorem. We shall now give an example in which the extremal function (8) does not degenerate into a function of the form s(z,
Izi < 1,
the maximum value of
If(z)1
d.
We know 1) that, for fixed z in
in the class
Tr is attained only by a fu"nc
tion of the form f(z) = s(z, t), where, for certain z, we have - 1 < t < 1. The function z 2/(1- z2)2f(z) belongs to T r whenever f(z) does. (This follows from the necessary and suHic ient condition for
f (z)
to belong to the class
T"
namely, that the inequality
ffi ( I /2 holds for class
1z
\
< 1.) Therefore, for
1z I
!(Z») >0
< 1, the minimum value of
If (z ) I in the
Tr is attained only by a function
f(z)
~
= (l-z2)2 S (z, t)
z (I +z)!
+ 2 (1 I-t
=-2- s (z, 1) Goluzin [195oa].
~
t) (l_z9)2 I
+t
1)+2- s (z, -1) E T"
The proof is analogous to that of Theorem 1.
THE SCIENTIFIC WORKS OF GENNADil MIHAILOVIC GOLUZIN (All in Russian) 1929
On certain inequalities dealing with functions executing a univalent con formal transformation of a disk, Mat. Sb. 36, 152-172.
1933
(with V. I. Krylov), A generalization of Carleman' s formula and its applica tion to analytic continuation of functions, Mat. Sb. 40, 144-149.
1934
The solution of the fundamental plane problems of mathematical physics for the case of Laplace's equation in multiply connected domains bounded by circles (the method of functional equations), Mat. Sb. 41, 246-276.
1954a Solution of the three-dimensional Dirichlet problem for Laplace's equation and for domains bounded by a finite number of spheres, Mit. Sb. 41, 277-283. 1935
On the theory of univalent conformal transformations, Mat. Sb. 42, 169-190.
1935a Solution of the plane problem of heat flow for multiply connected domains bounded by circles in the case of an isolating layer, Mat. Sb. 42, 191-198. 1935b On the majorization principle in the theory of functions, Mat. Sb. 42, 647-650. 1936
On distortion theorems in the theory of conformal mappings, Mat. Sb. 1 (43), 127-135.
1936a On the theory of conformal mappings, Mat. Sb. 1 (43), 273-282. 1936b On theorems of inverses in the theory of univalent functions, Mat. Sb. 1 (43), 293-296. 1937
On distortion theorems for conformal mapping of multiply connected do mains, Mat. Sb. 2 (44), 37-64.
1937a Some covering theorems for functions that are regular in a disk, Mat. Sb. 2 (44),617-619. 1937b Supplementary remarks to the article "On distortion theorems in the theory of conformal mappings", Mat. Sb. 2 (44), 685-688. 1937c On conformal mapping of doubly connected domains bounded by rectilinear and circular polygons in Conformal mapping of simply connected and mul tiply connected domains, ONTI, Moscow, pp. 90-97.
545
546
SCIENTIFIC WORKS OF GENNADII MIHAILOVIC GOLUZIN
1937d Conformal mapping of multiply connected domains onto a plane with cuts by the method of functional equations in Conformal mapping of simply connected and multiply connected domains, ONTI, Moscow, pp. 98-110. 1938
Some bounds on the coefficients of univalent functions, Mat. Sb. 3 (45), 321-330.
1938a On the continuitY.. method in the theory of conformal mappings of multiply connected domains, Mat. Sb. 4 (46), 3-8. 1939
Iterational processes for conformal mapping of multiply connected domains, Mat. Sb. 6 (48), 377-382.
1939a On limiting values of Cauchy integral, Leningrad State Univ. Ann. Math. Ser. 6, 43-47.
MR 2, 181.
1939b On complete systems of functions in the complex domain, Leningrad State Univ. Ann. Math. Ser. 6, 48-51. MR 2, 188. 1939c Zur Theorie der schlichten Funktionen, Mat. Sh. 6 (48), 383-388. MR 1, 308. 1939d Interior problems of the theory of univalent functions, Uspehi Mat. Nauk 6, 26-89.
MR 1, 49.
1940
Uber p-valente Funktionen, Mat. Sh. 8 (50), 277-284.
MR 2, 185.
1943
Uber Koeffizienten der schlichten Funktionen, Mat. Sb. 12 (54), 40-47.
v
v
19 (61), 183-202.
MR 8, 325.
Method of variations in the theory of conform representation. I, Mat. Sb.
19 (60, 203-236.
MR 8, 325.
Estimates for analytic functions with bounded mean of the modulus, Trudy
Mat. Inst. Stek1ov., 18, 1-87.
MR 8, 573.
The method of variations in conformal mapping, Leningrad. Nauen. Bju1!.
Univ. 9, 3-5.
Method of variations in the theory of conform representation. II, Mat. Sb.
21 (63), 83-117.
MR 9, 421.
Method of variations in the theory of conform representation. III, Mat. Sb.
21 (63), 119-132.
MR 9, 421.
Some covering theorems in the theory of analytic functions, Mat. Sb. 22
(64), 353-372.
MR 10, 241.
On the coefficients of univalent functions, Mat. Sb. 22 (64), 373-380.
MR 10, 186.
On distortion theorems and the coefficients of univalent functions, Mat. Sb.
23 (65) 353-360.
MR 10, 602.
On analytic functions with bounded mean modulus, Leningrad. Gos. Univ. Ucen. Zap. 111, 120-125.
1943b On the theory of the airfoil in the two-dimensional flow, Mat. Sh. 12 (54), 146-151. MR 5, 21.
Leningrad. Gos. Univ. Ueen. Zap. Ill, 126-134.
Some estimations of derivatives of bounded functions, Mat. Sb. 16 (58), 295-306.
1946
MR 7, 202.
On the theory of univalent functions, Mat. Sb. 18 (60), 167-179. MR 7,515.
1946a On the problem of Caratheodory-Fejer and similar problems, Mat. Sb. 18 (60), 213-226.
MR 8, 22.
1946b On some properties of polynomials, Mat. Sb. 18 (60), 227-236. MR 8, 22. 1946c On the distortion theorems for "schlicht" conform representation, Mat. Sb. 18 (60), 379-390. MR 8, 574. 1946d On the number of finite asymptotic values of integral functions of finite order, Mat. Sh. 18 (60), 391-396. MR 8, 23.
547
On distortion theorems and coefficients of univalent functions, Mat. Sb.
MR 5, 93. 1943a Zur Theorie der schlichten Funktionen, Mat. Sb. 12 (54), 48-55. MR 5, 93.
1945
"
SCIENTIFIC WORKS OF GENNADII MIHAILOVIC GOLUZIN
Variational derivation of a distortion theorem for univalent functions, Some inequalities for analytic functions, Izv. Akad. Nauk Kazah. SSR 60
Ser. Mat. Meh. 3, 101-105.
MR 13, 639.
Some questions of the theory of univalent functions, Trudy Mat. Inst. Stek
lov. 27.
MR 13, 123.
On mean values, Mat. Sb. 25 (67), 307-314; English transI., Amer. Math.
Soc. Transi. (1) 2 (1962), 275-283.
MR 11, 339.
Some estimates for bounded functions, Mat. Sb. 26 (68), 7-18. MR 11, 426.
On typically real functions, Mat. Sb. 27 (69), 201-218.
MR 12, 490.
On the theory of univalent functions, Mat. Sb. 28 (70), 351-358.
MR 13, 639. 1951a On the theory of univalent functions, Mat. Sb. 29 (71), 197-208. MR 13, 223.
548
v " SCIENTIFIC WORKS OF GENNADIIu MIHAILOVIC GOLUZIN
1951b On the majorants of subordinate analytic functions. I, Mat. Sb. 29 (71), 209-224. MR 13, 223. 1951c On subordinate univalent functions, Trudy Mat. Inst. Steklov. 38, 68-71. MR 13, 733. 1951d Variational method in conformal mapping. IV, Mat. Sb. 29 (71), 455-468. MR 13, 454.
BIBLIOGRAPHY L. AhHors
1930
1951e On the parametric representation of functions univalent in a ring, Mat. Sb. 29 (71), 469-476.
MR 13, 930.
1951£ On majorization of subordinate analytic functions. II, Mat. Sb. 29 (71), 593-602. MR 13, 454. 1951g On the problem of the coefficients of univalent functions, Dokl. Akad. Nauk SSSR 81, 721-723. 1952
MR 13, 546.
On a variational method in the theory of analytic functions, Leningrad. Gos. Univ. Ueen. Zap. 144, Ser. Mat. Nauk 23, 85-101; English trans!., Amer. Math. Soc. Transl. (2) 180%1), 1-14. MR 17, 1070; MR 23 #AI803.
1952a Geometric theory of functions of a complex variable, GITTL, Moscow. MR 15, 712.
UntersuchuT'gen zur Theorie der konformen Abbildungen und ganzen Funktionen, Acta Soc. Sci. Fenn. 1, no. 9, 1-40.
1938
An extension of Schwarz' lemma, Trans. Amer. Math. Soc. 43,359-364.
1947
Bounded analytic functions, Duke Math. J. 14, 1-11.
MR 9, 24.
L. Ahlfors and H. Grunsky,
1937
Uber die Blochsche Konstante, Math. Z. 42, 671-673.
Ju. E. Alenicyn 1947
On mean p-valent functions, Mat. Sb. 20 (62), 113-124. (Russian) MR 9,23.
1950
On functions p-valent in the mean, Mat. Sb. 27 (69), 285-296. (Russian)
MR 12, 491.
1950a On bounded functions in multiply connected regions, Dokl. Akad. Nauk SSSR 73, 245-248. (Russian) 1951
28 (70), 401-406. (Russian) 1961
MR 12, 401.
On the estimation of the coefficients of univalent functions, Mat. Sb. MR 13, 640.
An extension of the principle of subordination to multiply connected regions, Trudy Mat. Inst. Steklov. 60, 5-21; English transl., Amer. Math. Soc. Transl. (2) 43 (1964),281-297.
1964
MR 25 #1281.
Conformal mappings of a multiply connected domain onto many-sheeted canonical surfaces, Izv. Akad. Nauk SSSR Ser. Mat. 28, 607-644. (Russian)
MR 29 #1315.
I. E. Bazilevic
1936
Zum Koeffizientenproblem der schlichten Funktionen, Mat. Sb. 1 (43), 221-228.
1936a Sur les theoremes de Koebe-Bieberbach, 1951
Mat. Sb. 1 (43), 283-292.
On distortion theorems and coefficients in univalent functions, Mat. Sb. 28 (70), 147-164. (Russian)
MR 12, 600.
549
~-
-....:...
--0..s;--
_---
"---"'"
550
._,;;:-"------ _ _- "
_
"'-~-,---~
~~....-~
.......' ~ - =
~~~",-~~~-"~;""..,,-~*~~~~~BS?~;;/;;~~~~~~~;,;;,.-,,;£, -;:; 1;
S m for '(I> 1;
Izi < 1
and satisfy the inequality
Izi < 1.
the class of functions fez) that are regular in the disk
C
the class of functions fez) that are regular in the disk satisfy the inequality lR(f(z»
> 0 for
lzl < 1; Izi < 1 and
Izi < 1;
the class of functions fez) = z + c 2 z2 + ••• that are regular in the disk
Izi < 1; disk Izi < 1
and satisfy the inequality ~(f(z»~(z) > 0 for
the class of functions fez) that are regular in the
and that
satisfy the inequality fez I) fez 2) ,j 1 for arbitrary z I and z 2 in the disk
L
SO)
nSR.
Basic methods of the geometric theory of functions
,of the principle of areas leads to the proof of a "generalized theorem on areas". furthermore, such a theorem is the starting part for obtaining inequalities of :'various kinds. One of the first theorems of this nature was obtained by Lebedev i(and Milin [1951]. Lebedev [1961] gave a generalization and strengthening of this
~:result, which we shall now briefly expound. ':;" ~' Let ~ (00, al ,·· " a) denote the class of all systems lfk(z)l~ of functions ,~~ w = fk(z), for k = 0, 1,'" , n, that map the disk Izi < 1 conformally and univa
= 00 and fk(O) = a , for k Itk = 1, 2,'" , n, where aI' a ,··· ,a are fixed points. We denote by Bk(r), for n n 2 ~~k = 0, 1,"" n, the image of the disk Izi < r, where 0 < r ~ 1, under the function t: w = fk(k). We denote by B (r) the complement of the set U k= 0 Bk(r) with respect ~:'tO the extended w-plane. Suppose that a function Q (w) has a regular and single
'il:"lentlY onto pairwise disjoint domains such that fo(O)
lR
R
S~) is the class of function~ fez) €
565
proved the theorem on areas for p-valent functions. In more general cases, the use
0;
2
Izi < 1
BASIC METHODS
fashion by Prawitz [19271 and Grunsky [1939]. Later, Goluzin (see OJapter XI, §6)
> 1 onto
the class of functions fez) = az + a z2 +"', where a> 0, that are regular and univalent in the disk
T
_ =,;=;,~"."'"
...
By using the theorem on areas, Bieberbach proved the familiar theorems in the
onto a convex
SM (M ~ 1) the class of bounded functions fez) € 5: m
~',
)c1ass 5 and I (d. §4 of Chapter 11). The principle of areas was used in a finer
the class of functions in I
5(1)
~...:""," __"",.."""
used in its simplest form by Gronwall (d. the area theorem in §4 of Chapter II).
the class of functions fez) € 5 that map the disk
(m
.'_:
1°. The principle of areas. The principle of areas (area is nonnegative) was
onto a do
with convex complement;
I
,
of a complex variable
a domain whose complement is starlike about the point w
I
§ 1.
2' C 3' ... ;
domain;
I*
n SM' and
~-1f ..,
Other notation will be explained when introduced.
+...;
main that is starlike about the point w
S
functions f(z)€ S(k)
ak_/,k-l + a 2k _/,2k-1 the class of functions fez) = z + c 2 z2 + ... € 5 with real coefficients C
5*
1'1> 1;
that do not vanish for
1,2,"') the class of functions in IO that have an expansion of
the form F((} =
SR
§l.
7
Izi < 1;
i~valued derivative
in the domain B(r o )' where 0 < r < 1. The problem is to calcu o late the area a (r), where r 0 < r < 1, of the image of the domain B (r) under the lfunction ~ = Q (w):
a (r)
the class of functions f(z) that are regular in the disk satisfy the inequality fez I) fez )
~':'
f- -
Iz I < 1
1 for arbitrary z I and z 2 in that
disk;
'" '" R, L the subclasses of univalent functions in the classes Rand L respec
H I Q' (w) I~ dUl,
"here dw is the element of area. Consider some single-valued branch of the func tion Q (w). Then, in the annulus r 0 expansion
< Iz I < 1
Q (fdz)) = ~ ~~k)Z-·q
The classes of functions that are subclasses of several of these classes will
5 1, under the function w
=
v'=1
2
q Ibq 1
q=1
•
! f 1~' q\
q=1
v'=1
q
(;~,)q r·
(6)
tbe theorem is proved. Oae can find the conditions under which equality will hold
,t
of his theorems.
'V, ...
and satisfies the inequality If(z)1
Iz I < I
1 (1-1 Z.12)2 -
If' (Zl) 121M2 (I-If (Zl) 121M2)'
I
I}1/2 •
SUPPLEMENT
580
§2. UNIVALENT FUNCTIONS IN DISK ANNULl/S
If we now set z I
1
=
z 2' we obtain the well-known inequality (Alenicyn [1956])
61 {I, z} I
+
M 2 1f'(z)12 /A,f!
1
J'I~\
1m
~
I -'-I- I
~
12\2
(4)
,
where {f, z I is Schwarz's invariant. On the one hand, inequality (4) gives a bound
581
D of the functional 1=log(zA[,(z)I-Alf(z)A) (where A is a given real number) defined on the class S. He obtained a differential equation for functions fez) € S corresponding to nonsingular boundary points 10 of the domain D, that is, points for which there exists an a¢: D such that II -
I
1
ai, where
I € D, attains its mini
for Schwarz's invariant in the case of bounded univalent functions; on the other
mum with
hand, it sharpens for such functions the well-known inequality of the hyper
ary of D. This equation contains a parameter that is a root of a sixth-degree alge
bolic metric. Inequality (4) can also be obtained in an elementary way (see Lebe dev [1961]) from the well-known inequality of Nehari [1949] for the absolute value of Schwarz's invariant in the class S:
{I, z} 1 ~ - "- ,
,
The set of nonsingular points is everywhere dense on the bound
braic equation. As a special case, we obtain the complete solution of the problem . of the maximum and minimum of
~ (J)
= log 1 zA
f f(Z) 1 -AIfez )1..1.
tion, Lebedev [1955a] determined the range of values of the system 1a, log (f(z); z)J
'0'.'
in the class
We note that if a function f(z) that is regular in the disk
Iz 1
< 1 and satis
S(l)
of functions f(t;) = a( + a 2 (2 + ... that are regular, univalent, 1(I < 1 and also the range of values
and bounded in modulus by unity in the disk
of the functionals log (z['(z)lf(z)), 10g(az 2 ['(z)/[2(z)) and loga['(z) in the
fies the inequality
I{f, z}I~~ ,
class S(1)[lf(z)\l of functions in ,~.
1 and satisfy the condition F (00)
([(z), F(')) such that fez) For the class
n
n
Let mo denote the set of all functions fez) that are meromorphic and univa lent in the disk 1zj < 1 and satisfy the condition [(0) = O. Let 1Jll00 denote the set
(6)
W-ak lak=p
k=1
= [
n
I
v (z) of the equations
f-
k=1
F(,) for every z in Izi < 1 and every' in It::"! > 1.
Alenicyn [1956a] used Nehari's method to obtain the following
m C/. {
I
1
log
z2f' (z)f' (0)
i2 (z)
in
1'1
> 1, and arbitrary con
(C)} + 2C/.l~ log (1f-( ZF )(C) ) + C/.~ log F'F'(co) 2
~ - I C/.ll i log (1
I
z=e,p-~ (W, - a.) (n (W, - ak)ak)~,
00. Let 9J1 denote the set of ordered pairs
inequality for arbitrary z in Iz I < 1, arbitrary , stants a and a : l 2
partitions the w-plane into n simply connected domains. Extremal systems of functions are constituted only by the solutions w v
m,
=
-I
z I~) - I~ Ii log (1
-
l
I
C 12
).
(The values of the logarithms on the left-hand side are taken on suitable branches.)
k::j:.y
w,(O)=a" lel=l ('1=1,2, ''', n),
This result leads to inequalities involving various functionals both in the class of functions mentioned and for univalent mappings. In particular, Alenicyn
corresponding to the curves (6) that partition the plane into n simply connected domains. These solutions map the disk Izi
< 1 univalently onto the domains in
dicated.
obtained [1958] the following inequalities, which do not involve the derivatives of the functions in question: Let [l(z) and [/z) denote two functions that are meromorphic and have no
In the simplest case of n
=
2, Lebedev [1955b] posed and solved a more gen
eral problem. Specifically, let B denote a simply connected domain containing and a 2. If a k f- 00, for k = 1, 2, let w = [/z) denote a l function that satisfies the condition [k(O) = a k and that maps the disk Izi < 1 univalently and conformally onto a domain B k contained in B. If a 2 = 00, let
common values in the disk Izi
given distinct points a
[2(1;) denote a function satisfying the condition [/00) = 00 that maps the domain I" > 1 onto a domain B 2 contained in B. The domains B l and B 2 are disjoint. Let &, B (aI' a 2) denote the set of points M (x l' x), where x k = 1[~(0)1 if a k f- 00 and x 2 = 1[~(00)1 if a 2 = 00. Lebedev [1955b] obtained the result that, in the case in which B is the en tire plane (including w
=
00), the set &, B (0, 00) is the domain 0 < x 1 x 2 S 1,
< 1, and [/,) = Ot, for > 1 and 0 < t < 00, and the set &'B (aI' a 2) is the domain 0 < x l x 2 S lal-a212,
Xl> 0, with x l x 2 = 1 only when [1(') =
I"
t',
for It::"!
Xl> 0 with x l x 2 = la l - a212 only when [1(') = (alk l - a 2,)/(k l - ,), [2(') = (a l k 2 - a 2,)/(k 2 - ,), and Iklk21 = 1. Furthermore, by applying Goluzin's varia tional method, Lebedev determined the region &, B (aI' a 2 ) for cases when B is the plane with the point 00 excluded and when B is the disk Iwl < R. As a con sequence of these theorems, one can again obtain the results of Lavrent' ev [1934] and Kolbina [1952] (here B is the unextended plane) and those of Kufarev and Fales [1951] (here, B is a disk).
< 1. Suppose that [1(0) = 0 and [2(0) = 00. Then, < 1,
for arbitrary points z l' z 2 in the disk Izl ] log
(1 - j: ~::~) I~ -
Ij: ~::~ I~ I
Zl Z i
+
log (1 - I Zl
Ii) (1 - I z~ I~),
I/V(1 -I zll~) (1- I Z~ I~)·
These inequalities are sharp for arbitrary z 1 and z 2 such that Iz 11
=
I
z 21
=
1
with equality holding only in the case of univalent functions. These results lead immediately to a number of sharp inequalities for functions in the classes Rand L and functions associated with them (see Alenicyn [1956a, 1958]). For other results in this direction, in particular their extension to a cir cular annulus, see
§ 3,
subsection 2° of this supplement.
The same functional [1(zl)/[2(z2) in the class of functions indicated was studied by Lebedev [1957] by means of the variational method: for the class 1Jll, he found in explicit form the boundary of the range of the functional ~ =
[(zo)/F(,o) or, equivalently, of the functional ~= [(r)/F(p), where r= Izol and p = 1'0 I· The pairs of functions corresponding to given boundary points of this
range can be determined from the differential equations that we have obtained for
•._:
.• - -
-
.__...__.
-:_::-
.
_~_~
_..
_.
_._0_,:::.-
~.
__.
.....
~::_'"'_____
__.__
_:__.
__
_"'='
_"'0'_.
~
f.~
them. As a consequence, sharp bounds have been obtained for If(r)/ F (p)1 and Ilog (1 - fer)! F (p)) I, the radii of disks that prove to be the ranges of values of
Inequality (7) strengthens the previous result of Lebede; and Milin [19511: if
fez) €
R (or L), then
['(O)/F(p) and f(r)/F'(",,), the range of values of the system (If(r)l, 1/!F(p)\) in the class m, and also the range of values of the function fez) in the class R.
2"
i7' ~
o
The range of values of the system (1["(0)/['(0)1, I['(O)/F'(",,)\) in the class
mwas determined by Ulina [1960). maximizing the functional" J
=
If "(0)/[,(OW'-. 1['(0)/ F'(",,)!.B , a > 0,
f3 > 0
=
7fz, where 17f1
=
1.
In the problem of conformal radii of disjoint domains, Lebedev [19611 obtained
in
the following result:
Lebedev [19611 has obtained a number of results of extremely general nature in problems dealing with disjoint domains by beginning with the area principle.
a) of all systems Ifk(z)\~ of functions fk(z), for k = 0, 1," " n, that map the disk Izi < 1 conformally and univalently onto disjoint domains in such a way that fo (0) = "" and fk(O) = a k, k = 1" " , n, where a , " ' , an are fixed points. By using the generalized area theorem for He considered the class m("", a
1
functions in this class (see
§ 1,
1
n
IIlf';(O)IITkI2~ k=O
where f~(O)
m("", a 1 ,'" ,an)' in particu
2"
(8)
limz~o l/zfo(z). The cases in which equality holds in (8) were all
Inequality (8) generalizes to the case of complex Yk the inequality known A recent paper of Jenkins [19651 is closely related to this class of questions. eral inequality of the distortion-theorem type for the clas s 311, defined above, of
210
~ III (e ) I~ de· ~
Ifo (~i6)Ti de ~ 1,
i6
o
0
pairs of functions If(z), F«()l that depend on a large number of parameters. From that inequality, one can obtain a number of particular inequalities for the functions
fez) and F(,) and for functions in the classes Rand
with equality holding if and only if
I a I> I b I, II (z) =
"l =
ak - azl-2!R{TkTz},
In tbat paper, Jenkins started with the area principle and established a more gen
m("", 0), then
a fo(z)=-+b, z
j
l~k 0, does not ex ceed 21Tp. The function w = fez) is said to be p-valent in mean with respect to Riemann surface
with equality holding only for
f(z) =
fez) that is regular in the disk Iz\ < 1 is said to be p-valent in mean with respect to circumference in Izi < 1 if it maps the disk Izi < 1 onto a A function w
2"
i'IJz
p~l,
l"ll
=1.
area in the disk Izi < 1 if it maps the disk Izi < 1 onto a Riemann surface such
588
SUPPLEMENT
that the part of its area that lies over an arbitrary disk
Iwl :s p
theorem on the transfinite diameter (Theorem 3, Chapter VII,
Following Hayman, we shall say that a function f(z) that is regular in the
Izl
disk
f(z)
< 1 is weakly p-valent in that disk if, for every p> 0, the equation
w either has exactly p roots in \zl
=
< 1 for every w on the circle
or (2) has fewer than p roots in Iz I < 1 for some w on the circle
Iwl
=
Iwl
=
p
For weakly p-valent functions, Hayman [1951a, 1958] has obtained the follow
p. 1£ p is
ing generalization of Bieberbach's classical bounds on the modulus of a function and its derivative in the class S (see Chapter II, §4).
tion than the property of being p-valent in mean with respect to circumference, but
Let f(z)
= zP
+
being p-valent in mean with respect to area does not imply weak p-valence or vice
C + zP +1
+ ••• denote a function belonging to F'. Then,
p
i
P
ICP+ll~2p
versa. We denote by FO (respectively FDp or F'), where p is a positive integer, p p
Also, for
Izi
_z j( z) that are regular in the disk
P+ Cp+lZ P+l+
Iz\ < 1
rP cp+~z J+~+
... ,
II
and p-valent in mean with respect to circum !;Furthermore, w ~;
that disk.
in the disk
=
,
is regular in the disk
Izi < 1.
f(z) = w o + cpz P + •.. , where c p 1= 0 and P':::: 1
Let B = B
denote its range in the disk
Izi < 1.
f Suppose that the domain B* obtained from B by symmetrization about a straight
line or ray passing through the point
W
Suppose that a function w
o+
=
¢(z)
= W
o
lies in a simply connected domain B o'
1, the theorem was proved by Kobori and Abe [19591 This theorem has led to a number of distortion and covering theorems for regu lar functions, in particular, for functions in the classes FO and F'p (see, for ex p
,on~'.
Iwl < 4- P
exactly p times
Jz\ < 1. Equality holds in all these inequalities only for the function
fez) =
zP __ \21"
11
I e 1=1.
(10)
By combining the method of extremal metries that he developed and the re sults of the symmetrization method and by generalizing his own results for the class S [1953a], Jenkins obtained [1955, 1958], for f(z) € F? ' the least upper bound of If(r) I, where 0
< r 2 < 1, for a given value of If( - r 1) I, where r 1 is a fixed number in the interval 0 < r 1 < 1. In particular, we have the theorem: Suppose that f(z) € F?, that 0 < r1:S r 2 < 1, and that () is a real number. Then
1/(_rlei6)I+l/(r~eI6)1~(1~~r~)2
When p = 1, this theorem was obtained by Hayman [1951, 1958] by means of a combination of a bound that he obtained for the inner radius of the range of a regu
prP - 1 (l +r)
~\ I/(z) I ~"
11
f(z) assumes every value in the disk
A powerful tool in the theory of conformal mapping is the following symme
=
p (l +r)
(z) I ~ ~
trization principle. Suppose that a function w
fP - ~1/(z)l~ (l-r)~P'
-
ference (respectively p-valent in mean with respect to area, weakly p-valent) in
(9)
= r, where 0 < r < 1, we have the sharp inequalities
the class of functions of the form
Izi < 1
§ 3) for the case of
nonunivalent mappings 1) has also found applications in this class of equations.
a positive integer, the property of being weakly p-valent is a less stringent condi
disk
589
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
does not exceed TTpp2
If r 1
< r 2' equality holds only for f(z)
holds only for f(z)
=
=
+ (1+ r l)2' 1
z!(1 - e -je z)2. If r 1
= r 2'
equality
z!(1 ± e ie z)2.
As a consequence, we have the following inequality for a function f(z) € F?:
1/(-z)I+I/(z)l~
2r (l +r l ) 1 1 _ ..2\2'
Izl=r
(O 3. 3 This shows that the result Jenkins obtained for the class F~ does not hold for
class
the entire class F~. We note that the last assertion in Hayman's theorem has been carried over by D
Fp
•
With re
gard to this question, see also Jenkins's article [1957a]. We cite yet another example of the extension of familiar distortion theorems for univalent functions to nonunivalent mappings. We give the following definition: A family ~ of functions fez) is said to be
linearly invariant if it satisfies the following two conditions:
sup
! E~
ICgl
have been obtained. (Here a:::: 1. Furthermore, a is finite if and only if
53 is a
In particular, for a
=
2, we obtain the classical distortion theorems for univalent
functions (see Chapter II, §4). We now stop for certain results of the distortion-theorem type for doubly con nected domains.
with equality holding for p ~ 1 only in the case of the functions (10).
Garabedian and Royden [1954] to the case of functions in the class
(J.=
normal family, and a = 1 if and only if all functions in the family ~ are convex.)
cp+g+ p
1 C p +9 I ~
A
number of results with which we are familiar for univalent functions can be carried (12)
\c 3 1:S 3
1
Iz I < 1.
In analogy with Loewner's parametric representation of univalent functions in a disk, Komatu [19431 gave a parametric representation of functions w = fez) that are regular and univalent in the annulus I onto the domain
Iwl>
< Iz[ < R
and that map that annulus
1 with cut along a Jordan curve. Here, the circle
is mapped onto the circle
Iwl
Izi
=1
= I and f(l) = 1
Goluzin [1951e1 gave a simpler variation of the solution of this rroblem in a different form. Let K (m/M) denote the class of functions w = f(z) that are regu
< Iz I < M, where < r 1 < r 2 < M, map the
> 0,
lar and univalent in the annulus m
m
that annulus, and that, for m
circle
of the image of the circle
Izl
fez) € K(m/M) for which f(l)
= =
r 2'
that do not vanish in
Izl
= r 1 into the interior
Let K (m/M, 1/1) denote the class of functions
1 and m:s 1 :sM. Following the method indicated
in the article by Goluzin cited above, Li En-pir [1953] considered the problem of
592
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
SUPPLEMENT
onto the unit disk with cut along the radius connecting the origin with the point
generalizing Loewner's parametric representation for functions in the class
K(m/M, 1/1) that map the annulus m < Izi < M onto the domain obtained from the w-plane by deleting two Jordan curves that terminate respectively at the points 0 and
00.
- f(b). When b
metric representation that he obtained makes it possible to obtain a number of in equalities in the class K (m/M) and its subclasses with a complete clarification of the question of extremal functions. As an example, we give the following re =
fl(z) and w
map the annulus m
f 2(z) denote functions in the class K(m/M, 1/1) that
=
< Iz I < M onto the entire w-plane with cuts along a segment of
the real axis with endpoint at w
=
0 and along the ray lying on the real axis ex
f 2 (z)
on the nega
tive half. For the functions fl(z) and f 2 (z), we have explicit expressions (see subsection 4° of this supplement). We have the following theorem (Lebedev
§ 2,
For fez) € K(m/M, 1/1) and Izi
< qo < Po < 1.
where 0
Let
J
r: q 0 :s z :s Po'
denote the class of functions w = fez) that are
regular and univalent in B and map B into the unit disk in such a way that the point z = 0 is mapped into w = 0 and the circle boundary of the image. Let the condition
1['(0)\
Je
Iz I = 1
is mapped into the outer
denote the subclass of functions in
J
that satisfy
= c, where c is a fixed positive number.
Tamrazov [I96Sc] posed and solved the problem of determining the quantities
Pf= sUPz€r If(z)1 and qf = infZ€r If(z)l in the class Je for various positive values of c. Let f o(z) denote a function in the class J that maps B onto the unit disk with cut along a circular arc with center at the coordinate origin and
=
the classes
with equality holding in the first (second) inequality only for the function (f (z )).
(13)
f l (z)
2
Inequalities (13) had been obtained earlier [1953] by Li En-pir without the
< Iz I < 1
For c € (0, 1], the only extremal functions are functions of the form fez)
=
where
iEl
disk Iwl
= 1 and w = /l(z) = cz + .•• maps the disk Izi < 1 univalently onto the < 1 with cut along a segment of positive radius. For c = 1, this cut is =
1. For c € (1, co), the qualitative nature of the ex
tremal mappings is different. The solution of the problem in this case depends on
and that map that annulus onto a domain contained
extensive use of the theory of quadratic differentials, the general theorem on Jen
=
in such a way that the circle
Izl
=
1 is mapped onto the circle
Iwl = 1. Let F 0 denote the class of functions in F that satisfy the condition fez) i 0 in B. Duren and Schiffer [I962] developed the method of variations in the class F and applied it to solve certain extremal problems in that class. Duren [1963] ap plied that method to the problem of maximizing and minimizing the quantity If'(b) I
< b < lout
o
consists only of functions of the form
fez) that are regular and univalent
Let F denote the class of functions w
1. For the mean modulus of a function fez) E proved the inequality
21t
in o ~
I/(rei'f') I dep
5, he
< 1 r r + 0.55,
2
method of circular symmetrization (the solution of this problem also follows from
which deviates from the sharp inequality by no more than an additional term.
the familiar results of Teichmiiller [1938]), solution of the problem of minimizing
A number of res ults obtained by Bazilevic in 1958 (d. his article [I961])
belong to the same class of problems.
Pf in the class ~ required the application of variational methods (Tamrazov [1965c]). Specifically, we have the following theorem:
These results are closely connected with the problem of finding bounds for
Suppose that a function w = g (z) maps B univalently onto the entire w-plane with cuts
-00
< w :S - t (where t = t{r) > 0) and O:S w :S 1. In the class
have P/2- t with equality holding if and only if fez) where
If I =
=
g(fZ) or fez)
=
~, we
Icnl
1. Aleksandrov and Ceroikov [1963] first found the least upper bound of the curvature of the level curves in the class 5( k)*. For k = 1, this least upper bound
ctkR (r)
by the function f(z)
=
z/(l + zk)2!k; for
'0:::;' < 1,
0,
rJ.J«r) 1 with cut w? P.
= [*(z,
., boundary o[ the domain
values a s (r) and (3 s (r)
have been found such that every arc of a level curve L r is starlike for an arbi
=
that map B onto a set B f contained in the
1, and that map the circle Izi = 1 onto the circle
onto the domain
for all functions in the class S.
< r < 1,
Iwl>
~ that a [unction w
where (3 is independent of r, in which the arcs of the level curves are starlike
rJ.s
603
transfinite diameter, Kubo [1958] obtained a number of theorems for functions that
[1963] showed that there is no interior annulus of starlikeness, that is, no annulus of the form
or in the annulus
~""~---~---.-------
are regular or meromorphic in an annulus. We present one of these.
gard to the behavior of the level curves for small values of 1[(z)l, Stepanova
Popov [1965]: For every r such that tanh (TT/4)
_"c~==~-~="",·~~_,_,="'''''''''~~_~=~~~''~"_,,·,_c~~
By applying the ideas presented in Hayman's article [1951a] to the hyperbolic
Stepanova [1963] obtained analogous bounds for the annulus of starlikeness
Here, {3 s (r)
'-._C"_.'_ -- c,__ ,- c=, --," -,·,. ,c__,-.C"
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
SUPPLEMENT
and the annulus of convexity for the classes
!:".~,-"",.-,-,,-=-
in
n.
For an arbitrary disk
Iz -
trary ¢ in the interval 0:5 ¢
= a s = 0.109'" .
Stepanova [1965] established an analogous relationship between starlikeness
< p,
Lp(q:»=
Together with a more profound study of the geometrical properties of conformal
and let Zo denote a point
where p> 0, contained in
n
and for arbi
let us consider the set E of points z €
such that Iz - zol = r > p and arg(z - zo)
of arcs of the level curves and the values of 1f'(z)l. mapping by univalent functions, a number of results of the covering-theorem type
zol
< 2TT,
00
n denote an open
J\ dr r'
=
n
¢. Let us define
R(cp)=pexpLp(cp).
E
-Furthermore, for arbitrary n
=
2, 3," . , we define
have been established for general classes of functions. For functions that are regular in the disk Izi eralization of Koebe's theorem on Let [(z)
=
< 1,
n-I
~ L~n)(cp)=n "'"
there is the following gen
I
X' .
k=O
n-I
z + c 2 z 2 + ••• denote a member o[ lR and let d f denote the linear
measure o[ the set o[ all positive p [or which the circle
Iwl =
the image B o[ the disk Izi < 1 under the mapping w f equality holding only [or [unctions o[ the [orm
[(z). Then d ?
Z
f(z)='1_~Z)9
=
f
X',
II
RI/n
(q:>
+ 2:k) =
p exp
L~n) (q:».
Obviously, R(n) (¢) is independent of p. We shall refer to the transformation of the set
,lsl=I.
(q:» =
21tk) ' +n
k=O
with
This result was obtained by Hayman [1951a] by use of the geometric proper ties of the transfinite diameter.
R(n)
p is contained in
( Lp cP
n
and 0:5 ¢
into the set of points z such that z - Zo
< 2TT,
as the Sn transformation of the set
=
re i ¢, where 0:5 r
< R(n)(¢)
n with center at the point z O.
Under the Sn -transformation, an arbitrary domain is mapped either onto the .plane or onto a domain that is starlike about the point
Z0
and possesses for n? 2,
604
SUPPLEMENT
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
605
,
n-fold symmetry of rotation about that point. As Marcus has shown, such symme trization does not increase the interior radius of the domain. By using this last result, Marcus obtained the following theorem:
As a supplement to the results of Goluzin and Bermant regarding a covering performed by regular functions in a disk (see Chapter IV, §6), we present the fol lowing theorem (Jenkins [1951]).
Suppose that f(z) = z + c z2 ••. is a member of ~. Suppose that the S 2
n
transformation with center at the point w R(II)
(If)~
=0
-:/1 V 4'
is applied to the domain Br Then,
In connection with these results, we have the following theorem of Antonjuk
with equality holding for a function of the form
Ie I=
[1958] for functions that are regular in an annulus.
1.
Suppose that the function w
This theorem shows that at least one ray of an arbitrary system of n rays issuing from the point w
=
If f(z) E ~, then in the image B f of the disk Izl < 1 under the mapping = f(z) there exist n straight line segments issuing from w = 0 at equal angles
the sum of the lengths of which is arbitrarily close to n/r*, where p* = 1"* 2 is the area of the preimage of the star of the domain B!,
0 ~ If< 21t,
j(z)= (l-:ZII)I/II ,
w
0 at equal angles to each other intersects the domain
and satisfies the conditions
=
f(z) is regular in the annulus B: 1_ < Izi
If(z)1 2:
1 and
,
ICpl~IC;1
with equality holding only for a function of the form f(z)
o = 0 in are their multiplicities. Mitjuk [1965d] W
also obtained an analogous strengthening of Hayman's theorem on ~ that we have given and of Marcus's theorem.
In connection with the above-mentioned result of Kubo [1958], we mention that Mitjuk [I965b] obtained a more general covering theorem for functions that are regular in an annulus. Specifically, he considered a broader class of functions and studied the multiplicity of the covering produced by them. We mention also that Mitjuk [1965c, d] extended Hayman's theorem of subsection 1° to the case of functions that are weakly p-valent in an annulus. For functions of this last class, he obtained a theorem on the linear measure of the covering of arcs of a circle, thus generalizing the theorem of Kubo [1954] mentioned above.
3°. Bounds on the coefficients of univalent functions. The problem of the coefficients in the expansions of univalent functions has played a significant role in recent years in the general turn taken by the geometric theory of functions. As we know, for functions f(z) mining for every n
Ic 2' •••
, Cn
I
=
z + c 2z2 + ••. in the class S, it consists in deter
2. 2 the ranges of values of the system of coefficients
of functions in that class. A particular case of this problem is the
problem of finding sharp bounds for the coefficients. Success in this search would lead to proof or refutation of a well-known conjecture of Bieberbach.
In recent years, a number of problems associated with this class of questions have been solved. Schaeffer and Spencer [1950] used their own form of the variational method to determine explicitly the range of values V 3 of the system of coefficients
Ie 2' c 3 1
606
SUPPLEMENT
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
in the class S. Although V 3 is a body in 4-dimensional real Euclidean space, the
2
IlXnl~ n+l'
property of symmetry with respect to rotation admits a simple description of this set if we determine its three-dimensional cross-sections for which, for example, = 0 (c = a + ib , where a k and b k are real). The boundary of the k k k 3 range of values of each of the systems {a 2, a , b J, {a 2, b 2, a31 is expressed in 3 3 explicit form in terms of elementary functions.
b 2 = 0 or b
In the class 1 of functions F(,)
=
X + e -6. of
,-3
l
a
3
~
1
This proves that the bound found by Goluzin [1949c] for the coefficient
for odd functions in the class
1 is sharp throughout that class.
In 1955, Garabedian and Schiffer [1955a] first proved the validity of Bieber bach's conjecture for the fourth coefficient. This proof uses both the method of variations and Loewner's parametric method. The ranges of values of the systems of initial coefficients in classes of
(1)
Let M denote the class of functions f(z) meromorphic and univalent in the disk
Izi < 1
" function fez)
that satisfy the conditions f(o) =
z + 1~-2 J-
C .
J
zi E M, and if
'+ a o + a 1'-1 +"', Garabedian and
Schiffer [1955] obtained by the method of variations the sharp inequality
607
C .
= =
0 and ['(0) 0 for J'
J
=
1. If
n>
2, if the
< (n + 1)/2, then
2 n-l'
ICnl~
(2)
Equality holds in (1) and (2) only for the functions F(,; n + 1, a) and
F- 1 (z-l; n - 1, a) respectively, where F(,; n, a)
=,
(1 + e ina ,-n)2/n.
Inequalities (1) and (2) were obtained by Goluzin for the classes
1 0 and S
t respectively as special cases of results that are valid for p-valent functions (see Chapter XI, § 6). If n > 0 and 00
bounded functions in the classes
S(k)
and
l(k)
[1957] by means of Loewner's parametric representation method. Thus, in the class S<j;) of functions fez) = z + c k +1z k +1 + c2k+1z2k+1 +"', where c Vk +1 = a vk +1 + ib vk +1 for k = 1, 2,"" Bazilevic determined in explicit form the range of values of the system
{Ie k +1 1, Ie 2k +1 1J and,
b 2k +1 = 0, the range of values of the system
under the assumption that b k +1
la k +1' a2k +11
=
and he exhibited all
their boundary functions. Bazilevic posed and solved the same problems for the
1 1 onto a domain bounded by a Jordan curve. 2
I
amnxmxfl ~
equality converges, where the coefficients a
2: 1. In connection with this, we note that
Izl < 1
< 1 to be univalent in
be satisfied for every sequence
Clunie [1959a] constructed along the lines of Littlewood's well-known construc tion a function in the class
Izi
~ ..i..J Im,n=
and satisfies for all n the differ
AN~ULUS
I ca -
: c:
g
:
c~ I ~ 4,
I~ ~a. -V4 -
g 1 Cg I ,
(9)
what modified form. The proof rests on Grunsky's familiar condition [1939] for univalence (see also Chapter IV,
§ 2).
We present this result in the special case
of functions that are regular in the disk
1
z 1 < 1: I)
1) The article by Garabedian, Ross and Schiffer [1965], gives an elementary proof of this theorem that is based on the area method.
Without loss of generality, we can assume in the problem that we are con sidering that c 4 2: o. Then, from (8) we have
1) Inequality (8) can also be obtained directly from inequality (12) of Chapter IV, §2, with n = 3 and the same values for x l' Xz, and x3'
610
SUPPLEMENT
611
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS "
Ct ~ 21/12 + ; +ffi {2 (C2 -I) (CS -
:
The Grunsky-Nehari inequalities, which give necessary and sufficient condi
5 12 C~ _/2C2}'
C~) +
tions for univalence and boundedness of a function in the disk
and, in view of (9), this leads to the inequality
Ct~21/12+ ~
+
-:a IC2-/IV4-lc212+ffi{152c~-f2c2}'
If we now set
1=2xe
_i.'L
=
O~x~
rp,
2
1,
I
in the interval 0
fl) (b
I)
of functions f(z)
< b I < 1,
Isin(3¢/2)1,
.
For example, for 0
by a function w
bIZ + b 2 z 2 + ... € 5
fl)
2: O. Schiffer and Tammi [1965] examined this
< b i < 1/11,
fo(z) that maps the disk
=
the maximum value of b 4 is attained
Izi < 1
onto the unit disk with cut
< 1, the f I (z) of the disk Iz \ < 1 onto
lying on the negative radius and only by such a function; for 19/34:S b I maximum value of b 4 is attained by a mapping w
=
the unit disk with three radial cuts of equal length making equal angles each with
and, consequently,
the next. In these statements, the constant 1/11 cannot be replaced with a larger
2 + 8X2- -14XS - (8 X2- -4X 3)Y 2+ -----=-X 8 '/-1 Ct
SUPPLEMENT
612
613
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
2,~:2Ick - k1 ::; f m is satisfied for f(z) = z + c 2 z 2 + ... € 5, then ~c2m:S 2m with equality holding only for the function [*(z). 2
For odd functions, it is well known that
\ b~n_l I ~ A,
Furthermore, the problem of global proof of Bieberbach's conjecture for the where
in five independent variables. In connection with this last, we point out that
a uniform inequality of the form
::;
6 in the class
5 for
> 1,
A is an absolute constant. It has also been shown that A > 1. Furthermore, Ib2 n -11 ::; 1 for n 2: N is impos sible (see Chapter IV, § 7). The exact value of the constant A is unknown. By using the bound for
sixth coefficient is reduced to the problem of proving a trigonometric inequality Bombieri [1963] had alre~dy proved the inequality ~c 6
n
Iz I ::; r < 1
values of ~c 2 sufficiently close to 2 and Ozawa [1965] used Grunsky's condi
the area of the image of the disk
tions (7) to prove this inequality under the assumption that 0 S c 2 ::; 2.
(see the article by Lebedev and Milin [1951]), Gong Sheng (Kung Sun) [1955] ob
With regard to the question of finding bounds for the coefficients for all
n> 4,
tained the inequality
I bin_ 1 \ < 2.55, n ~ 1.
the best result in this direction that has so far been obtained is the inequality found by Milin [19651:
under a function in the clas s S( 2)
Let us denote by 5 (a) and 5( 2) (a) respectively the subclasses of functions
Ic n I < 1.243n,
> 4.
This result was obtained by investigation of univalent functions based on use of
in th~ classes Sand 5(2) for which (10) holds with fixed a = a, where 0 0 as r --'> 1, holds for coefficient5 in the expansion (11) of the function 4J (z).
(2)
614
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
SUPPLEMENT
Recently, Hayman [1963] obtained a bound on the growth of adjacent coeffi
It follows, in particular, from (12) that co
V k '-
cients that is asymptotically sharp not only for the class 5 but also for the class F~.
Ilk-Te-JkOO 1 . l'i =T1n-;-, 1 1
If a function f(z)
=
k=1
and f(e iflO ), then, for arbitrary real A and
T,
In the particular case in which
the inequality
such that
1
+i
k=1
where g (T) tion $(z) the disk
=
-->
10k Ii r ik ~ (
)41
:2~1~ I
(l -.:
_1)2
=
1:S n V+I
(l
+ g(r)),
:S K,
= 0 for a sequence n = n v' for
V=
1, 2,' .. ,
we have
n ~ no'
I Cn I.:;;;;;; AK,
(15)
Inequality (15) cannot be improved as regards order of dependence on K, as is shown by the functions CX)
1 + G1z + G 2z 2 +"', which is regular in
fa (z) =
Izi < 1.
V
Z
1
'l_
~~~ A
I
_2
sinn6
= '- sin 6
n Z •
n=1 'V
'V
.
mean square modulus in the classes 5(a) and 5(2)(a), the inequality giving a lower bound on the distance between adjacent coefficients in the expansion of the
In particular, if
mean in a joint paper by Bazilevic and Lebedev [1966] with the aid of methods ex
TTlK, where K is even, then for n = K(m + Y2L
C
n
=
0 for n
inequalities regarding regular functions, we also use the following finer result: Let f(z) denote a function belonging to F~. Let z 1 with
T
1
KITT
V
function f 2(z) = f(z 2) and various other inequalities characterizing the spread of the coefficients of univalent functions.
1
zll'
(16)
2
where A is an absolute constant.
The problem of investigating the interrelationships between the order of growth of sequences of coefficients is related in a natural way to the question of the asymptotic behavior of the coefficients in the expansions of univalent func tions. The first systematic results in this direction belong to Goluzin (Chapter
IV, §8).
For f(z) E 5, the inequality obtained from (16) by repl~cing its left-hand member with If(z2)1 was obtained by Biernacki [1956] from an inequality of Golu zin [1946e, 1949c]. F or functions in the class 5, this proof of inequality (14) is simplified only
Goluzin also examined the problem of finding bounds for the growth of adjacent coefficients in the expansions of functions in the class
5 (ibid.). For functions in
5*, he obtained [1946e, 1949cJ the inequality f
to an inconsiderable extent. In connection with this, it is appropriate to cite Hay man's assertion that it is difficult to prove a result of an asymptotic nature for univalent functions by a method that could not at the same time be applied to func tions that are p-valent in mean.
I Cn+l I-I cn II < A
A is an absolute constant less than 100), which is sharp with regard
the order of dependence on n.
nv
(13)
Special cases of inequality (13) are the inequalities giving a bound on the
(where
-
Cn
2
0 as T --> 1, holds for the coefficients in the expansion of the func
[(f(z)jz)(1 + e- iflO z)2A]T
(14)
where A is an absolute constant.
Milin [1965] is used, this theorem leads to the following result:
S(a)
z + 2~=2 C nzn E F~, then
ricn+1 1- I cn II ~ A, n ~ 1,
When an inequality for regular functions that was obtained by Lebedev and
If f(z) E
615
to
Lucas (d. Hayman [1965]) extended the result given above to functions that are p-valent in mean:
616
SUPPLEMENT
If f(z) =
+
Co
C 1Z
+ ."
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
is regular and p-valent in mean with respect to area
obtained the following result:
in the disk Izi < 1, then
!lcn :_ll-\c n l[=O(n
gp
If f(z) = z +
-'.l), p~l,
1~=z cnz n € 5
(17)
I
1 /
(1S)
Inequality (I7) cannot be improved and, for p < 1, it is no longer true. For
Furthermore, for a
00.
Let us choose n k
< 1. Then, the function
00
!(z)=z+ ~
n'/lInkZnk
k=1
is bounded, univalent, and starlike in the disk
Izl
< 1
For functions f(z) = z + 1~=zcnzn E 5, for which If(z)\ < M/(I-lzl)a. in the disk Izi < 1, we have, for the coefficients
form the set V*n in (n - I)-dimensional complex Euclidean space consisting of all points (c z' .•.
shown by the following example:
a> Y2, the well-known sharp bound on the increase in
I cn I< M· A (Cl)n- I +
ll ,
where A (a) depends only on a. (See, for example, Hayman [195S] 3.3; an analo gous inequality is valid for p-valent functions.) For a ~
Y2,
Pommerenke [1961/62]
, c)
that are put in correspondence with at least one function
f(z) = z + czz Z + ... + c zn + ... € 5 * . Here, V * is determined from its twon
dimensional cross section
*
en'
for which c z'
n
... , C n -1
are fixed. If (c z,' .. ,c n -1)
* -1' then en* is a disk whose radius does not exceed is an interior point of Vn 2/(n - 1). This is a sharp bound since it is atrained when c Z = c 3 = •.. = c n-I = 0 by a function of the form f(z) = Z/(1 - (Zn-I)2/(n-l), where If\ = 1. In particular, the boundaries of the cross-sections C*Z and C*3 are determined by the equations C
z = 2e
i8
, and c 3 =3c;/4+ e i8 (4-l c l)z/4. z
In the class K (see p.619) of functions f(z) = z + to convex in the disk Izl < 1, Bieberbach's conjecture
C
zzz + ••• that are close
Ie n l ~n is valid, as Reade has shown [1955]. For functions F (() = ( + a
o
+ a / ( + ...
618
1(1 > 1, Pommerenke [1962] has
that are meromorphic and close to convex in
The structural formula for the class C also leads easily to integral represen
ob
tations of the classes
tained
I I= (Xn
that is, functions [(z) that are regular in the disk
z + I~=2cnzn E K, then
and regular and convex in
t. 0
°2 m{l J C
is sharp with respect to the order of dependence on n. In particular, for all func tions in 5* except the functions
z
=
I cn +ll-1 cn \ =
o(n > 0 0
This class was introduced by Kaplan [19521.
Izi < 1
and [(z) is close to convex
+zr (Z)}dO>-1t
f' (z)
61.
°
< 01 < 2 :::; 17 and z
ie
re , where r < 1. The class of close-to-convex functions coincides (see Lewandowski [1958,1960])
for arbitrary 01 and 02 such that -17
Izi < 1
with the class of functions that map the disk
(l-ei&I Z) (l-i0 2z) ,
where 01 and 02 are real, there exists a 0
and satisfy there the
and
n"?= 1.
Pommerenke also indicated the form of all functions for which this inequality
f(z)=
Izi < 1.
Every close-to-convex function is univalent in if and only if f'(z)
II cn +ll-1 cn II < : e~,
Izi < 1
condition ~1f'(z)/g'(Z)j > 0, where g(z) is some function that is independent of [
cients for univalent functions, Pommerenke [1963] proved the following theorem: =
5* (see Chapter IX, §9) and -- S. By this procedure, one can
find an integral representation for the class of functions that are close to convex,
0 ( ~ )•
Supplementary to the results given above, on the growth of adjacent coeffi
I[ [(z)
619
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
SUPPLEMENT
=
onto domains linearly acces
sible from the outside, which Biernacki [19361 had first investigated. We denote by K the class of functions [(z)
such that
and close to convex in the disk
(~5)'
=
z + c 2z2 + ••• that are regular
Izl < 1.
It is obvious from the definition of the class K that 5* eKe S.
4°. Structural formulas for various classes of analytic functions. A large num
Ju. D. Maksimov [1955] introduced into consideration the class C(p) of (
ber of results have been obtained for classes of functions that can be represented
locally-convex functions and the class S/p) of (-locally-starlike functions [(z)'
by some "structural formula" or other, in particular, by aStieltjes integral. It is
with [(0)
well known that the class C of functions [(z) the disk
Izl < 1
=
1+
C 1z
+ ••• that are regular in
O2
sented by the Stieltjes integral
"
~
(1)
=
Robertson [1935] and Goluzin [1950a] used this formula to obtain an integral repre sentation for the class T (see Chapter XI, §9). Beginning with this representa tion, Goluzin [1950a1 obtained in the class T a number of distortion and rotation r
< 1.
JC\rn {1
dO> - e,
(z) } _ + zj" f' (z) dO -
2p 1t
-It
O2
f' (z)
on the
"
}dO> - e ,
01
1.
On the basis of a simple connection between functions in the classes C and T,
=
+ I' (z)
~ m{Z~'(~i
where /1(t) is a nondecreasing function in [-17,17] such that /1(17) - /1(-17)
Izi
z 1 < 1 and that satisfy the conditions
for [(z) E C(p), and
eit+z e' -z
- . t - dfl- (t).
theorems and also bounds on the mean moduli of [(z) and
I
11
'zr (Z)}
&1
-11
circle
0, that are regular in the disk
C { ~ m1
and satisfy the condition ~[(z) > 0 in that disk can be repre
f(z)=
=
=
I(z)
-It
for [(z) E S(p) in the annulus 0:::; p
z
Cm{ zl' (z) }dO = 2p1t
J
< Izi < 1, where
p depends on the function
·e re ' ,-17 < 01 < 02 :::; 17, and (> O. Let C(p, q) and S(p, q) denote subclasses of functions in C(p) and S(p)
that are of the form
fez) =
zq
+
q c q +1Z +t
+...
.__
~
~:_ ~ _
!!!
•
_~.:_
_'!.- .-
620
--
'~.:.::
·§rtiW:~"'..ii"
;OJ!!! - - - .:....
_~
:n.;,'J§.!'_~"'"
_if,-;;:;;,-"i2;~,,;;""
ii5lGiliiiiZi'""o"'t-"~c:;;;:';';~'~-.>"--::'.~_ ~""'
*,,~~i~~,"::7i"""-:'~
""o.'
~~~~-""-li::;"
-=
-"..._~:.,-
621
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
SUPPLEMENT
-~""':1'~~~--
These classes contain many well-known classes of functions. In particular,
Loewner-Kufarev equation in quadrature in a special case, Bazilevic [1964] gen
C1/1, 1) is the class K defined above, the class S11(I) contains functions rhat
eralized his previous results by proving the following theorem: If two functions po(z) and Pl(z) are regular in the disk Izi
are starlike with respect to direction (see Robertson [1936]), and the classes
positive real parts in that disk, then the function
Co(p) and So(p) were examined by Goodman [1950l. Ju. D. Maksimov [1955, 1955a] has obtain~d an integral representation of the
z
ing max~lo. ""2 ,=1
and which assumes a given value w at a given point Zo
Cz (l -
e-ICXz)1 •
the role of Koebe's function for a disk. For example, the function (5) with a
=
and only such a function attains extrema of If(z)\ for all z such that
q < Izi < 1 in the class U*(q, 1) (and in the broader class of functions that are
< Izi < 1
[see subsection 10 of §2]) and also extrema
of argf(z) for arbitrary z in the annulus q
f.
0 in the disk
Izl
< l.
In the article by Andreeva, Lebedev and Stovbun [1961], problems are posed
The function (5) plays in many problems a role for an annulus analogous to
univalent in the annulus q
whose Maclaurin-series expansions begin
Co
fWlction
f3 ± IT
IT,
with a given polynomial
fez)
Uiiiii.J.iii!
< Izi < 1 in the class U*(q, 1) (see
Dunducenko [1956]). We mention that Zmorovic [1956] obtained an integral representation for yet more general classes of fWlctions that are regular in an annulus with real part
and solved regarding functions in the class C (r) for which the problem formulated above, of Monastyrskil, is a special case. In that article, Lebedev solves the fol lowing general problem: Let M denote the class of functions w
Izl < 1
=
fez) that are regular in the disk
and let G denote an arbitrary simply connected domain in the w-plane that
does not include the point
00
and some other given point. We shall say that the
function fez) € M is subject to the domain G if, for arbitrary z in the disk Izi
< 1, the point
w = fez) € G. We denote by Me the class of functions that are
subject to the domain G. Let z 0
=
0 and let z k (k
=
1", • , m) denote distinct
624 points in the domain 0
< Izi < I,
numbers, and let w~v) (k We use the notation x
=
=
let a k (k = 0, I,""
0, I,""
m; v = I, 2,""
Alenicyn [1962] has obtained results of a very general nature regarding the
m) denote given natural a ) denote give~ numbers. k
-
-
(...,(1)
",.,(110) '4'0' ••• , UVO I " "
",.,(1)
",.,(l1
U'lm' ••• , U'lmm
))
in it by a sum of Stieltjes integrals, and he obtained their boundary functions:
•
Let G denote a finitely connected bounded circular domain in the z-plane and
Let M (x, a, X) denote the class of functions f(z) € M that satisfy the conditions ('I
11)Ij (f, MG) of functionals
(f, MG) of functionals (7) defined on the class MG ' that is, find the set of points X in n-dimensional complex Euclidean space R n , where n = 2:=1 a k , such that, for every point X = X (w~ v» in this set there exists a function f(z) €
and A2 such that A'
~ n=-oo
The following problem is solved: find the range of values of the system
that, for any two numbers \
(z) =
gk(Z, t)= ~
a~.k(t)(z-zyt,v=l,... ,s, k=l, ... ,m,tEra,bJ,
n=-oo
are valid in a neighborhood of the point Zv in the first case or in a circular annu lus with center at z v in the second. Furthermore, let
I C l' •••
,
C N I denote an
arbitrary fixed system of coefficients C~V) considered as a point
P [n
=
R N and let P[gk(z,t)] I, 2,'" , m) analogous to it defined in accordance with (8). Since the range of values of the system IC 1'" • , CNI in the class F is the range of values of the functional P[J] on that class, this range is the convex IC 1 , · · · ,
CNI inN-dimensional complex Euclidean space
denote the point of RN (k
=
envelope of the union of the sets U a:S t:S b P [g k (z, t)] for k = 1,' •• ,m. As an application in the class T of the functions f(z) = z + c 2z2 + •.• , one can find the range of values n(T) of the system l[(zo)' c21, where Zo is a nonzero fixed point of the disk Iz I F or ~ z 0
f-
< I:
0, the range of values of n (T) is a body in three-dimensional real
space bounded by two conical surfaces
Iw where f(zo) = W
(1
= Xl
~o ZO)I l~ ~ (Zo + zip) ---:- (X~ -+- 2) ~ { (I ~o Zg)1 W}, +
iX
2
and c = x • 2
3
._,~.".,;_~~,__ ~.:!t:,._~"'.'f~~~~~~.~"'-"!,~:""":5i.'!'J.E,~,n,,,"Z,,~~
.,=""",",=~-,=,,--=-...;;:::.-=-",,~~~~~""""=~===
626
-,>,~~~;P~'S;...s;;"':"1'i,",,-"="""·'~·~~f~;.~&;'~:,,3.c':"'?""-'_"'}"'i'f~'",.... _,,_ .•. ..:."',: 0,
Iwl
0, this function attains its greatest lower bound or its least upper bound in any disk \w the quantity
J = extr
tence of functions f(z) in the class T that have given initial segments of their Taylor expansions at different given points of the disk
Iz I < 1
C
N
°
= c 2' where - 2 < c 2 < 2, we obtain the result:
in the disk
1+
f(Z)=~A l+ze-i~k
function f(z) ranges over the class T (c 2)' When we look at the intersection of 3
°
f (z) =
note the subclass of functions in C of the form
T with fixed coefficient c . Let D.(T(c 2)) denote the range of values of f(zo) 2 under the condition that Zo is a fixed point in the domain < I Zo 1 < 1 and the the domain D. (T) with the plane x
627
§2. UNIVALENT FUNCTIONS IN DISK AND ANNULUS
SUPPLEMENT
wol S R
on the boundary of that disk. Then,
••• ,
us1 l'VS1 t; ... ;
llop,
'Vop ,
.,., llspp,
not vanish in the disk Izi
'Vspp )
z
The only boundary functions in the range of values of the functional 1 fined on the clas sCare functions in the clas seN' where N 2p - 1.
(n
de
:s s 1 + ... + S P +
=
l([) = J (f(z), fez), ['(z), ['(z» defined on the class C are functions in the class eN for N:s 2.
In particular, for functions fez) € H S that do
they obtained sharp bounds on I['(z)! for given
re i8 , where r < 1. The inequalities giving these bounds depend on the rela
[1962] by a different procedure. These include a number of results for the class H S [1961, 1962, 19631. defined on the class A M (see Chapter IX, § 3) of functions fez) are regular in the disk
o :S r < 1
c 1 z + ..• that
~ log+ If(re i9 ) \d9,;::;;:;M. o
metric representation of these classes of functions. Gel'fer [1965] applied Golu
§ 3.
and y is real).
+ C 1z + ••• that are regular in the disk Iz I < 1 and satisfy there the conditions If(z)\ < 1 and f(z) t- O. Using its simple connection with the class C, they arrived at a Dumber of inequalities in that class. Gel' fer and Kresnjakova [1965] con sidered the class H s (where [; > 0) of functions fez) = Co + C 1 z + ... that are
=
and that satisfy for arbitrary r in the interval
2"
zin's variational formulas in the class T to the problem of finding the maximum in Galperin [1965] obtained an integral representation of the class of functions fez) '"
Iz I < 1
the condition
We mention some of the other recent results that are based directly on a para
Iz I < 1
< 1,
Gel' fer [1966] used the same method to find the extrema of certain functionals
It follows, in particular, that the only boundary functions of the functional
that class of ~ Ie it' zf'(z)j f(z)l for fixed z and y (where
is a given entire func
tionships between 0 and r. One of them had already been found by Kresnjakova
denote a weakly differentiable function defined on the class C.
Co
< 1.
depending on its zeros in 1t;;"1
p). Let
< 1, (w)
tion, and B (t;;") is an appropriate multiplier in the parametric representation of f(t;;")
1
0
•
Functions that are analytic in multiply connected domains
Conformal mappings of multiply connected domains onto canonical sur
faces. In Chapter V we presented the basic extremal properties of conformal map pings of a multiply connected domain onto certain very simple canonical domains: onto the plane with parallel rectilinear cuts, onto the plane with cuts along arcs of logarithmic spirals all of the same inclination, and so forth. For the theory of
630
§3. FUNCTIONS ANALYTIC IN MUL TIPLY CONNECTED DOMAINS
SUPPLEMENT
631
been considered earlier (see Garabedian and Schiffer [1949]).
functions that are analytic in multiply connected domains, it will be of interest to answer the question about the existence and extremal properties of conformal map
Thus the function P (z) defined in terms of a suitable function S{z; (, a),
pings of a multiply connected domain onto canonical domains of a more complicated
and only this function minimizes the area of the image of the domain B in the class of all functions that are regular in that domain and that have fixed initial segments
type. We mention certain results associated with this question.
-in their Taylor expansions at given points of B (segments such that this class
Grunsky's article [1939] initiated the study of conformal mappings of a given multiply connected domain onto many-sheeted canonical surfaces. In that article,
does not include a constant). This provides a generalization both of the theorem
he studied a function that is regular in a multiply connected domain except at a
of minimization of area that was given in Chapter V, §4 and of Bergman's theorem
pole of multiplicity m and that maps that domain onto a Riemann surface with
[1950], which he obtained by a different method and in a different form, on the
parallel rectilinear cuts. For mappings that are defined on a multiply connected
minimization of area in the class of all functions that are regular in the domain B
domain and that possess the geometric property mentioned, though with singular
with fixed initial segment of their Taylor expansions at a given point of the domain. We define the outer area A (f) of a function f{z) in a domain B as the quan
parts of the mapping functions of a more general type, Alenicyn [1964] obtained
tity (finite or not)
the following results by the method of contour integration: Let B denote a bounded finitely connected domain in the z-plane with bound ary C consisting of simple closed analytic curves Cl'
... ,
Cm' Let (
l' ••• ,
where 5 ~ 1, denote arbitrary distinct points in the domain B. Let a oI)., a 1 ,}. , ' " ..• , a .. , where j = 1,' .. , 5, denote arbitrary coefficients (not all zero) such P )'1
that
A if) =
(5'
S (z;
~,
/ (z)j' (z)dz,
c(,)
interior of the domain B as v
--7 00.
Then the function
Q(z), and it alone, maxi
:, mizes the external area in the class of all functions of the form f(z) s
Il) =
~i ~
, where the C( v) are the boundaries of the domains B( V), which approximate the i'
Define
lim ...... 00
i :
I;=l aO,i = O.
-
!
}=1
p.
[1:
k=1
(z
~'C~)k + Gto.
j
log (z -
~j)]
=
S(z; (, a) +
g (z)' where S{z; (, a) is a fixed function of the singularities and g (z) is an
arbitrary function that is regular in the domain B. The properties of the function e{z) can be used to obtain theorems
(a function of the singularities). Under these conditions, we have the theorem: For an arbitrary given angle
e in the interval
-17/2
< e ~ 17/2 and an arbitrary
given function of the singularities S (z; (, a), there exists a function, unique up to an additive constant, e(z)
=
S(z; (, a) + Fiz) where Fe(z) is regularl) in
the closure of B, that possesses the following property: on every boundary curve
of the distortion-theorem type for functions that are meromorphic and p
valent in mean with respect to area in a multiply connected domain. A very
special case of the corresponding theorem on the range of values of a certain
functional defined on the class of functions that are meromorphic and p-valent in
mean with respect to area in a given multiply connected domain is the following:
CJ.I- of the domain B, every branch of the function e-iee(z) has a constant im aginary part.
In the class of all functions f(z) regular in the domain Izi
The functions
of a fixed point z 0 IeI
P (z) = 2 [ID~ (z) 2
I
n (z)l, where n
=
1, 2,' .• , is constructed satisfying the following conditions: The func
tions el>n(z) are regular in the closure of B and they have an expansion of the form eI> n (z) == !,OOk_- Jl. a n kZ -k , where a n n > a' for n == 1 , 2 , " ' , in a neighborhood of
634 z
=
SUPPLEMENT
635
§3. FUNCTIONS ANALYTIC IN MULTIPLY CONNECTED DOMAINS
00; the derivatives ¢'n (z) constitute an orthonormal system in the domain B.
~~~
tpk (z)
tp~ (z) do =
8k n> k, n = 1, 2, •.• ,
B
that is complete in the class L~ (B) of all functions that are regular and square integrable with a unique inverse in B. Let 1
[v (z) into the boundary component of the domain D v that is retained as the domains D v' for v = 1, 2, are broadened to a system of disjoint domains. Let =
[(z, 0, C) denote a function that maps the domain B univalently onto the unit
C) =
-
-all - a1
It)
=
= -1P Ell FII (Z,
ZII
)
1.
over (see Alenicyn [1966]) the results of Bergman and Schiffer [1951] mentioned above to the case of several functions that are univalent and have no common values in a multiply connected domain. Specifically, we have the theorem:
Let B denote a bounded and finitely connected domain in the z-plane bounded by closed analytic Jordan curves. Let [k(z), [or k
I 1: {!
with equality holding only in the case of functions defined by the equations
I, (z) I, (z)
'=
I122 (z) (z) -- all a1
)
Z, Zl,
= 1,'"
, n denote [unctions
denote arbitrary constants [or /l = 1," . , Nand k = 1, ... , n. Then, N n
Q21 2 1f' (0, 0, C1)f' (0, 0, Cg) \'
II (z) - a 1 = pf (z, 0, C1), 11 (z) -a.
1
common values in that domain. Let (kIJ- denote arbitrary points in 8 and let akIJ
0, where v'= 1, 2. Then
If; (O)f~ (0) I:E;; I Ql
It)
F (
that are regular and univalent in the domain 8 and suppose that they have no
disk with concentric circular cuts such that C v is mapped into the unit circle and [(0, 0,
and
pEl
Starting with an examination of suitable DiricWet integrals, one can carry
2
denote a sequence of functions that map B univa
lendy and conformally onto disjoint domains D v' Define a v = [)O) for v'= 1,2. Suppose that the boundary component C v of the domain B is mapped by the func tion w
°
Z2) F~ (Zg, ZIl)
deleted. Equality holds in the second inequality only
00
11 (z) - a1 = I 1 (Z ) - a2
(Chapter IV, §4) on univalent conformal mappings of a disk onto two disjoint domains.
finite numbers. Let
639
p..
>=1
~
2
2-f (z, 0, CIl),
IJ.kp.lJ.k. [Uk
k=I
+ 1 (~kp.' ~kv)] 0
V 1: IJ.kl'-~k.Ko (Ckl'-' ~k»'
I] (~jr,) Ii. (Ckl') } I IJ.jl'lJ.". (f· (C· )-h (~k»lI ~ ~
+-; l~j 3 -
2) the inequality
arbitrary point z 0
=
there exist functions satis
IF(zo)!;
in the annulus 0
< Izl ~
3-
is, but, for an
'1/ 8, there exist functions satisfying the
Thus, Biernacki's problems posed in Theorems 6 and 7 have obtained a
=
and satisfy the conditions f(o)
1. Suppose that F (z) is univalent in
1
=
F (0)
0,
=
If(z)! :::; F (z) in Izi < 1. Find the greatest radius R of a disk Izi < R in which Lewandowski [1961] showed that such a number
R exists and that 0.21 < R ~
0.29' • '. Under the additional assumption that F (z) is starlike in
!z I
oint in B to one of the functions in the class P Tt' then
1['(z)1
(~ F'(z, z)) for z €
z \ only for functions
B with equality holding at the point z
=
with equality holding (when Zo 1= a) only for fez)
o is a simple uncovered point of the function fez) in the domain B if the function log [fez) - w o] is regular in that domain. Then: If the function fez) is regular in a domain B and satisfies the conditions =
=
1, where a € B, and if If(z)!
MF'(a, a) (x> 0) in the disk
< M (MF'(a, a) > 0, then
Iwl < Me -x
When the domain B is the unit disk, this theorem yields Landau's familiar theorem [1929].
fH(z, 1; ao(zo)' a), where
In the case of the annulus B q ,we have (Alenicyn [1958]) the following rela
§4. We shall now elaborate on this question, using the remaining results of that paper and Alenicyn's paper [19611. Let B q denote a circular annulus q
n
H(z; a),
~=-argH'(a,
1;-
~, a).
Starring with Robinson's inequality (3), which is obviously a generalization of the first of the inequalities in the Schwarz lemma (see Chapter VII,
§ 1)
to the
case of a circular annulus, we can obtain theorems of a general nature that en able us to carry over to the case of a circular annulus certain distortion theorems and its simplest consequences. For brevity, we write H (z, 1; a (z), a) = H (zL o Let 9Jl* denote a class of functions g (z) that are meromorphic in the disk
Izi
< 1 and let 9Jl a (B) denote the class of all functions fez) that are meromorphic
in the domain B and that are subordinate in that domain about a point a in it to
Certain results from Robinson's paper [1943] were presented in Chapter XI,
< Izi < 1,
one of the functions in the class 9Jl*. We indicate this relationship between these two classes by writing '!!J!a(B) -< 9Jl*" Let us suppose that (w) is a real function
(lzl)
Let M
a 1 € Bq
Under these conditions, we have the theorem:
there exists a unique function H (z; a 1 ) that is regular in the closure of
the annulus B q , that is univalent in B q itself, and that satisfies the conditions
1, H(l;at)=l. H(at;at)=O,IH(z;at)I,Ii! 1_ - =Iatl, q . IH(z; at)l\ z /_1By using the corresponding results of Robinson [1943], we find, after some easy
m*. Izl O. We know from Robinson's paper [1943] that, for arbitrary given
transformations,
=
(3)
with which we are familiar for a circle. We present one of these general theorems
but it is not always true
that this function does not have simple uncovered points in a larger disk.
,
zo,
I f I = l-
F (z, a)= ei@z-tH(z; -
W
0, ['(a)
arg
1!(zo)I~IH(zo, I; adzo), a)1
Suppose that a function f(z) is regular in a domain B. We shall say that a
=
(q/\al) e i
in the annulus B q and an arbitrary function f(z) in the class ~(1)(B q' a), we have
!(z)=!(zt)+eF(z, Zt) (!(zt)+eF'(z, Zt», lel=I.
f(a)
=
tionship between the functions exhibited by AhHors and Robinson:
of the form
x -1 sinh x
we set ao(zo)
z-\ H(z; ao(zo»H(z; a). Then, for an arbitrary given point Zo
and map it onto surfaces the area of each of which (with the multiplicity
that is regular in a domain B is subordinate (univalently subordinate) in B about
point
=
,
1
(M/1z!)) denote a decreasing (increasing) function of
Izi
for
If one of the inequalities
M t (I
Z
l)~cD(g(z» ~M2 ([ z I),
I Z 1< 1,
holds in the class '!!J! , then the corresponding inlquality
*
(4)
~~
--
-
~.
648
_
..
~
;"";;;;"""~"~.-".~
Mt(IH(z)I)~(j(z))~M2(IH(z)I),zEB q ,
m*,
Here, equality holds at a point z
gfH(z)fH(z, 1; a O(zl)' a)), where
for fez)
=
1Zo I < 1
the function g
1fl
=
problem of finding =
1 and for a point
z 1 (~a) only Zo
with
(z) is extremal for the corresponding inequality (4).
the constant e
that annulus about a point a in it to one of the functions in the class S,
1-11~\ 12'
"_ ",_____
;;;.;:;;.;",,~~
___,_
._....._.. "'
Z
sUPf€iR(l)(Bl'.r cf(z)w(z)dz!
649
either is identically equal to
(where a is a real constant) or maps B onto the m-fold covered Nk
r
.
2
=1
The problem considered in this theorem generalizes, in particular, the problem
For an arbitrary function fez) that is regular in the annulus B q and subordinate in
I H (z) I
ia
J
From this theorem it follows, in particular, that
_I
,,"c,"
unit disk, where m ~ n. If w(z) = ~k=1 ~j=1 Il kj /(z - z)J, where a EB and the k
+ ~~ N k'
Ilk. are constants, then n S m -
zo
Ij(z) I ~ (1
";'~~~~
§3. FUNCTIONS ANALYTIC IN MUL TIPL Y CONNECTED DOMAINS
SUPPLEMENT
holds in the clas s ~I)B q)