The Bieberbach Conjecture

Chapter 7

The Bieberbach Conjecture By the Riemann Mapping Theorem (Chapter 1, Section 5), given a simply-connected region D in the plane which has at least two boundary points, there is an univalent function G mapping D onto B(0, 1), and further, given z0 E D, we can require G(zo) = 0 and G'(zo) > 0. Under these conditions G is unique. Alternatively, if instead we are willing to map D univalently by a function F onto a disk B(0, p), and require F'(zo) = 1, as well as F(zo) = O, then p is uniquely determined, and is the mapping radius of D, which has already been used (Theorem 6.5.2). Suppose now we consider the function r inverse to F on B(0, p); furthermore, suppose we consider D such that 0 E D, and z0 = 0 (this involves no loss of generality). Then -- - 1 , a n d r r is univalent on B(0, p); r B(O,p) -~ D, r - 0 and r F'(r 1 1 is the unique such function, for a given D. Finally, clearly the map f ( z ) - ~r is univalent on B(0, 1), f(0) = 0, and f'(0) = r = 1. Thus, after normalizing, we are led to study the class of functions f univalent on B(0, 1) with f(0) = 0, f'(0) = 1. Implicitly study of such functions is, by the Riemann mapping theorem, the study of simply connected regions in the plane. On the other hand, one way to study functions is to study their coefficients. The Bieberbach conjecture is a conjecture about the coefficients of normalized univalent functions on B(0, 1). A tremendous amount of work was done on it and related problems before it was proved by Louis de Branges in 1988. Again, this chapter can only attempt to be an introduction to the subject. Fortunately for the interested, there are books covering the subject in depth; in particular, Pommerenke, Univalent Functions [203], and Milin, Univalent Functions and Orthonormal Systems [157], as well as useful survey articles by Duren [62], and earlier, by Hayman [106], though this is now a little outdated. In Section 1 below, several distortion theorems other than those immediately relevant to the Bieberbach conjecture are discussed largely because of their intrinsic interest. The reader may wish to look up or try to discover other results which may be obtained by these essentially elementary methods. The Bieberbach conjecture itself is stated in Section 2, and a proof is given following

257

258

7. Bieberbach Conjecture

the ideas of de Branges and others.

Elementary Area and Distortion Theorems

7.1

Definition 1.1. S denotes the class of functions f univalent in B(0, 1) such that f (0) - O, f ' (0) - 1. It is often useful also to consider simply-connected regions G in Coo, where cc E G. We assume G has at least two boundary points. Picking a point ~ E C ~ - G , the map 1__ transforms G into a simply-connected region properly contained in C. Thus ~h~e Riemann Mapping Theorem shows that there is a unique function r - bz + ~~k=O bkz-k univalent in G and mapping G onto C ~ - B(0, 1) (with c~ going onto co). This leads to the following normalized definition. D e f i n i t i o n 1.2. E denotes the class of all functions g(z) - z + Ek=O bkz - k univalent in Coo - B(0, 1), except for the simple pole at c~. N o t e 1.1

9

If f E S, then g(z) - ~

1

__

--

z-l+Ek=2a~z-k -- +E.=la-+l 1

__

oo

z

1

"'

oo

z_

n

"--

Z - -

+ . . . ~ and so g(z) E E. Furthermore, since f E S ~ g(z) # 0 for z E Coo - B(0 ~ 1) z Conversely, if 9 E E, and ~ is in the complement of the image of Coo - B(O, 1) 1 1 z under 9, then f ( z ) - 9(~),r -----I;+bo-r bk~k -- l+(bo-r ~z" = Z § ( ( -- b o ) z 2 + . . . , and so f E S. (The choice of a ( is, of course, necessary for f to be analytic in B(0, 1)). Note, however, that every function in S omits the value c~, whereas functions in E need not omit any value. "

T h e o r e m 1.1. Suppose f ( z ) - z + ~'~v~176 under f has area 7r(1 + ~ - - 2 v[av[2) 9

~ E S, then the image of B(O, 1)

Proof. This is just the well-known "Area Theorem", which we have already used before (e.g. Theorem 1.5.3) specialized to B(0, 1) and al - 1 and which is proved in the Appendix. [--]

T h e o r e m 1.2. Suppose g(z) - z + ~v~176 bvz -v E E, then the area of the complement of the image of Ccr 1) under g is 7r 1 -

ub~

.

~,--1

Proof. Consider the circle C(0, r), r > 1 under 9, C(0, r) is mapped onto a simple closed analytic Jordan curve C with the equation w - w(8) - g(re i~ (r is fixed). The area J of the Jordan interior of (7 is (with w = u + iv) given by j

=

=

/o 2-

=

/02. (w0, +2 w/0,)

dO.

259

7.1. E l e m e n t a r y Area and Distortion Theorems

Now w' (0) - ~ ( r e i~ + E , ~ o b~ r - " e - ~ i ~ get J -

/o2 (r ei~176 -~

+ e

+ ~ Z

- irei~ - i E~~176 vb~r - ~ e -~i~ and so we

r - ~ (b~e-~i~ + ~ e ~ i ~

u--O

_

)

"

dO.

_

v--1

On multiplying and integrating termwise, all terms which contain a term e ki~ k r 0 will vanish, which gives J -

fo 2€ r 2 2

1 -l (bl _ ~ ) _ 1 cr 4 (bl - bl) + -~ ~ Z 2ulb~12r- 2~dO

= 7r r 2 -

vlb~12r -2v v--1

)

v--1

(1.1)

.

But J _> 0, and so, Z~=~ v[b,[2r -2" _< r2; thus m

v--1

and so, letting first r -+ 1, and then m -+ oc, we see that ~ =co 1 vIb~[2 converges; whence, letting r ---> 1 in (1.1), we get, as claimed, J - 7r

1-

vlbv[ 2

.

v--1

Vl

As an immediate corollary of J _> 0, we also get T h e o r e m 1.3. I / g ( z ) - z + y~,,~=ob~z -~ e E,

then EvC~=lb,[bv[ 2 <_ 1;

in particular,

Ibll <_ 1.

N o t e 1.2. Theorems 1.2 and 1.3 are due to Gronwall [89]. Note further that if then not only is [51[<1 but if [ 5 1 1 - 1 t h e n g ( z ) ~ ~ ~ z + bo + e i~ -1, for some real 0. It is remarkable that these non-obvious analytic facts follow from the simple geometric proposition that area is non-negative. One may also note that unless g(z) = z + bo, the area in question in Theorem 1.2 is always < ~r. g ( z ) = z + ~ ~ 1 7 6 v--O b ~ z _ V E Z

E x a m p l e 1.1. Let F ( z ) - En~176 n z n =

z

(1.z)2

'

Then F is analytic in B(0, 1). An easy computation shows that if F ( z ) - F(w), and z r w, then z w - 1, hence F is univalent in B(0, 1), and so F E S. F maps

260

7. Bieberbach Conjecture

C(0, 1) onto the straight line segment L - { z ' z r e a l , - co _< z _< - ~1} described twice, and the image of B(0, 1) under F is C - L, which has infinite area. The corresponding function in E, G(z) - ~ 1 = z - 2 + - ~ 1 maps Coo - B(0, 1) onto Coo - { z ' z r e a l , - 4 _< z < 0}. z3

E x a m p l e 1.2. Suppose F ( z ) - z - -y. F is clearly analytic in B(0, 1); and an easy computation shows that if F ( z ) - F ( w ) and z ~ w, then z 2 + zw + w 2 - 4, whence F is univalent in B(0, 1), and so F E S. The image of B(0, 1) under F is as in Diagram 7.1. z3

Image of C(0,1) under z - T

f// /

I [

t

-1 I f

~\ ] X

\

I

~,il

Diagram 7.1 1 and has area r(1 + 3(~) 2) - ~ . A corresponding function in P~ is Z3

oo

c~

G ( z ) - z2 - 1/3 : Z 3 - ~ z i - 2 ~ - z + Z 3 - ~ z - ( 2 ~ - 1 ) u=O

u--1

So the area of the complement of the image of C - B(0, 1) under G is

(

1 - ~(2rr=l

1)(3-~)~

) ( 1 2r 1) 2,3 - ~

1 - 5

3~

r=l

"

7.1. Elementary Area and Distortion Theorems

261

See Diagram 7.2. z3

Image of C(0, 1) under ~2_! 3

-I

Diagram 7.2 E x a m p l e 1.3. Let f ( z ) - ez 2 + (l__Zz)2, ~ > 0. f is clearly analytic in B(0,1). However, f is not univalent in B(0, 1). For, suppose it were in S, then the function g(z)

-

1 f ( ;1)

=

z 2 ( z - 1) 2 z 3 + e (Z - 1) 2

would be in E; however, g(~)

-

~

;

1

+ ~

+ E~%~ ~z-~

= z -

( 2 + ~) +

(1 + 4e + e2)

z

+...

contradicting Theorem 1.3. A direct proof that f is not univalent seems considerably more involved. E x a m p l e 1.4 . Let f(z) _ azk + (i_z)2, z k an integer > 3, c~ any non-zero real number, then f is clearly analytic in B(0, 1); however, f is not univalent in B(0, 1). For, suppose f E S; then 1

g(z)-

f(1)

I

k-I

v

(a-'t- k)

~

v

would be in E. But g(z) - z - 2 + 71 _ o ~ z k - 2 + . . . , which contradicts Theorem 1.3. Again, a direct proof that f is not univalent seems considerably more complicated. Theorem 1.3 already indicates that Theorems 1.1 and 1.2 may have non-obvious analytic consequences; another important instance, which explains Example 1.3, is (20

b'

T h e o r e m 1.4. Suppose f(z) - z + ~ = 2 a~z E S, then [a21 <_ 2.

7. Bieberbach Conjecture

262

Proof. The idea of the proof is to make a suitable transformation of f, pass to the corresponding function in E, and use Theorem 1.3 to obtain information about f. Thus we note that CX3

S(z )

+ Z

-

'

v=2

and let 1

F ( z ) - ( f ( z 2 ) ) 8 9- z + 7a

3

z +...

F(z) has the following properties: (i) All powers appearing in the expansion (1.2) are odd. (If an even power appeared, on squaring, we should get the contradiction that f(z 2) contained an odd power of z in its power series expansion.) Hence F is an odd function. (ii) F is analytic in B(0, 1). For ](0) = 0 and f is univalent in B(0, 1), so f ( z 2) takes the value 0 in B(0, 1) only at z = 0, where it has a double zero, and it follows that (f(z2)) 89- F(z) is analytic (and F(0) - 0 ) . (iii) F is, in fact, univalent in B(0, 1). For if

F(zl) = F(z2),

(zl,z2 e B(O, 1))

(1.3)

then f ( z 2) - f(z22), and so (since f is univalent in B(0, 1)) zl = +z2. If zl - -z2, then since by (i) F is odd, F(zl) = - F ( z 2 ) , and so by (1.2), F(Zl) = 0, whence f(z~) - O, and so zl - 0 - z 2 . By (i)-(iii) above, F is an odd function E S. Let r ~ be the corresponding function in E. Then r

-

1 1 ~_?_+ z + z" ""

1

= z -

~a2 + . . . . z

It now follows from Theorem 1.3, that l y1 a 21 _< 1, and so la21 _< 2. N o t e 1.3. If, for example, instead of F we use H(z) - (f(z2)) in B(0, 1) and

H(z)

_

1

4

z + ha2z +

...

1

; H ( 71)

1a

"- Z - -

~

--~

2

[--I

89then H is univalent

+""

and Theorem 1.3 (which shows that Ibll 2 +21b212 < 1)now gives la21 < ~2' a poorer result than Theorem 1.4.

7.1. Elementary Area and Distortion Theorems Similarly,

1

263

1 + . . . and we get no further information from Theorem = z - ~a2

1.3, if we try to use this. Indeed, Example 1.1 shows that Theorem 1.4 cannot be improved. An immediate consequence of Theorem 1.4, that was already proved in a quite different way (Note 6.5.4) is T h e o r e m 1.5. Suppose f E S. If B is the boundary o] the image of B(0, 1) under f , then every point o] B has a distance of at least 88 from the origin, (i.e. the image 1 contains the disk B(0, ~)); and 88 is the best possible such constant. Proo]. Suppose for z E B(0, 1), f ( z ) 7t C (there is some such C by Liouville's Theorem). Then C ~ 0 (f(0) - 0) and so fl(z) - cel(z) - f ( z ) is univalent in B(0, 1) c~ a~ ZU , But letting f ( z ) - z + ~-'~=2

C f (zl C

-

f(z)

--- Cz + Ca2z2 + "" -- z + la2 + --C1/ z2 C

-

z -

a2z 2

.

.

.

, 9 9

.

and s o f l E S . It now follows from Theorem 1.4 that la2 + -~1 _< 2. Hence I-~1 <- 2 + [a21, and 1 which proves the so again by Theorem 1.4 (since f E S), I~1 _ 4, or ICI >_ ~, theorem. Example 1.1 again shows the constant ~1 is best possible. D Theorem 1.4, taken with Note 1.3, leads to the question: for what functions f E S does [a2[ = 2. This is answered by the following T h e o r e m 1.6. If f c S ; f ( z ) - z + ~ - ~~=2 a~z~ , and [a21- 2, then f ( z ) for some real ~. Proof. If f E S F(z) - (f(z2)) 1/2 and r

z (l_eiOz)2,

- F--~), 1 then the proof of Theorem

1.4 shows that r E E where r - z - ~-~ z +~v~=2b~z -v. By Theorem 1.3, 1 2 ~la21 + ~ ~ =oo2 u l b ~ l 2 -< 1. So if la21 - 2, all the b~ - 0 for u >_ 2, and hence r

- z-

e'~ whence F ( z ) -7-,

z and 1,e~e:2

SO f ( z )

-- ( l _ e , O z ) 2

[::l

N o t e 1.4. Theorem 1.6 explains Example 1.4 above. Theorem 1.4 was first proved by Bieberbach who also discovered Theorems 1.2 and 1.3 independently of Gronwall [90]. The existence of a constant M such that B(0, M) is taken on by every f E S was first noted by Koebe [134], and the exact value M - 1 found by Bieberbach (implicitly) op. cir. and Faber [69]. Because of theorems like Theorems 1.4 1.6, the function (1-z)~ is often called the "Koebe function" and the functions (1_~,0~)~ are called "rotations of the Koebe function . Another proof of Theorem 1.5 (different from the ones above and in Note 6.5.4) can be found in Ahlfors [3] Conformal Invariants, p. 29. An important result further justifying a special denotation for these functions is the following "distortion theorem".

7. Bieberbach Conjecture

264 T h e o r e m 1.7. If f e S, then for all z E B(O, 1), Izl < If(z)l (1 + Izl) 2 1-izl < If'(z)l (1 + Izl) 3 -

< Izl - (1 -Izl)

2

(a)

< 1 + Izl - (1 - Izl) 3 '

(b)

and equality on any side only holds for some rotation of the Koebe function, i.e. for z 0 real. a function of the form (i-e'Oz)2, Proof. The proof again depends on Theorem 1.3, except that this time we first use a MSbius transformation in order to obtain nontrivial information. Given a fixed point zo E B(0, 1), the function g(~) - f ( 1~+zo + ~ ) i s a composition of univalent functions and so univalent in B(0, 1). Expanding in a power series around 0, we have g(r

g"(0);: 2 +""

= g(0) + g'(0)r +

(1.4)

= f(zo)+ f'(zo)(1 -Izol:)~+ + {f"(zo)(1 -Izo12)

2

- 2f'(zo)(1 - I z o l ~ ) ~ } ~

~2 - +...,

and so

h(r =

g(r

-

f(zo)

Writing

(1.5)

~S.

f'(zo)(1 -Izol ~)

h(~) - ( + ~2( 2 + . . . ,

we have from (1.4) and (1.5) that ]~2

and from Theorem 1.4 that zo E B(0, 1),

f"(zo)zo

2lzol 2

f'(zo)

1 - Izol~

--

f"(zo)(1 -I~ol

1~21 _ _

~)

2f'(Zo)

--

_

Z0

2zo 2; hence, multiplying by l_lzol2 we have for any

4]zol 1 -]zol 2 "

(1.6)

Since zo was an arbitrary point of B(0, 1) and [Re w[ < [wl, it follows from the triangle inequality that for all ( E B(0, 1) 2]r I-I~I 2

< Re (~f"(r ) < 21~12+41~I -

f'(r

-

1-I~I 2

(1.7)

7.1. Elementary Area and Distortion Theorems

265

However, writing ~ = re i~

d Log f, (re ie) ) - r d Re Log f' (re i~ r-~r

=Re

Re ( f , ( ( )

(1.8)

d

- r~rr Log lf'(r

9

Hence, (1.7) becomes 2r - 4 d 1 - r 2 -< ~rr Log If'(r

<

2r+4 1 - r2 "

Integrating throughout from 0 to Iz], we get (since f'(0) - 1 by hypothesis)

Log

( l_,z, ) (1 + Izl) 3

< Log If'(z)l < Log

(

(1 - l z l ) 3

)

or, by exponentiating, 1 -Izl < If'(z)l < 1 + Izl for z e B ( 0 1) ( 1 + Izl) 3 - (1 - Izl) 3' ' '

(1.9)

which proves (b). Now, for z E B(0, 1), integrating f'(() along the straight line from 0 to z gives (since f(0) - 0) by (1.9), If(z)l -

L Izl

z f' (()d(

(1 -Izl)

~

<_

Lz I/'(c:)Id(Ic:l) <_L z (11-I~I) + Ir

3 d(lr

(1.10)

"

On the other hand, let F be the curve with endpoints 0 and z such that f(F) is the straight line from 0 to f(z). Since f is univalent F does not cross itself and is rectifiable. Also, if ( - re i~ then an easy computation shows that Id(] >__dl(I, and so we have, putting w - f((), [f(z)l >

if(z) JO

Idwi=

J~FIf/ (()lfd~i> JfF

(1 + ICI) 3 dlC] -

tf'(C)ldlCi

(1.11)

(1 + ]z]) 2 "

(1.10) and (1.11) together prove (a). If equality holds in one side of (b), say for Zl # 0, then the proof shows that equality must hold in the corresponding side of (1.7) for all ( with 0 < i(I _< iZll 9 Thus, taking ( - rzl,0 < r _< 1, and dividing both sides of (1.7) by r, and then letting r --+ 0, we get (since f'(0) - 1), that equality must hold on one

266

7. Bieberbach Conjecture

side of-41z~l < R~(z~I"(O)) < 41~1. But if Re(zlf"(O)) - • I, then 41z~l < Izll If"(0)l and so IS"(0)l _> 4. But then by Theorem 1.4, I S " ( 0 ) l - 4, and so by z for some real Theorem 1.6, f ( z ) = (l+e,0z)2 Since (a) follows from (b) by integration, similar remarks hold for it. r3 N o t e 1.5. Theorem 1.7 also shows immediately that for f E S, and z E B(0, 1), l+lzl -< i(z) - 1-1zl ; however, one can do better by noting that if f E S, then also for the function h defined by equation (1.5) of the above proof, h E S, so, for all ~ e B(0, 1),

(1

Ir +

Given any point z0 E B(0, 1), let r h(-zo)

_

Ir

_< (1 -Ir 2 "

Ir = < lh(r

fg(-zo)-,

-z0, then by equation (1.5) above,

f(zo)

(z0)(1 -Izol ~)

since g ( - z o ) = f(O) = O. So we have

Izol(1 -Izol) 1 + Izol

-f(zo) $'(zo)(1-1zol2) '

f(zo) _< Izol(1 + Izol) <- S'(zo) 1 -Izol

or (replacing zo by z)

zf'(z) 1 -Izl i + Izl <- Siz)

l+lzl for a l l z E B ( 0 1 -I~l'

1).

Here again equality can only hold for a rotation of the Koebe Function. One may also note that taking z - re i~ , in the left side of Theorem 1.7 (a), and letting r ~ 1-, one obtains another proof of Theorem 1.5. Theorem 1.7 is also due to Bieberbach in Gb'ttingen Nachrichten [24], as well as Pick and Plemelj (see p. 946, footnote 2 of Bieberbach's paper) who earlier assumed [a21 _< 2, first proved by Bieberbach. As another example of the same sort of arguments we have been using, we have the following somewhat surprising T h e o r e m 1.8. Suppose r is univalent in $ - Coo - {z 9z <0}; that is the plane with origin and the negative real axis deleted. Then Ir is non-increasing on the positive real axis. Furthermore, if Ir takes on the same value at two different points of the positive real axis, then r - Az + B, where A, B are constants. Proo]. The theorem can be proved by mapping B(0, 1) onto the slit region S and applying Theorem 1.4.

(

Precisely, any function of the form c~ ~ maps B(O, 1) onto $, and is univalent in B(O, 1).

where c~ is a positive real number,

7.1. Elementary Area and Distortion Theorems Hence F(z)

-

267

r (o~ (~_-=~z)2) is univalent in B(0, 1). Furthermore, expanding

F in a power series about 0,

F(z) - F(0)+ F'(O)z + = r

+ 4~r

F"(0) 2 z2

+...

+ (S~2r

(1.12) + S~r

2

+...

and so if f(z) --

4~r

then f E S, and by Theorem 1.4,

~r r

+ 1 _< 1, for all positive real a .

Hence, Re

r

"(~)) <- 0 '

and so, since a > 0,

d Log 1r ~
but this means, by Theorem 1.6, that f(z) ilZz)2 (the Koebe function itself is the only rotation of the Koebe function for which the second coefficient has argument 0). Hence -

r

a

1

z

r

l+z c~ ~ - z

(l-z) 2 §162

or

2 =a

l+z 1 Z

where a and b are constants, which proves the result.

+b, [3

N o t e 1.6. Trivial examples of Theorem 1.8 are r = Log z and r - z 1/m, m a positive integer. By applying a preliminary rotation first, Theorem 1.8 can be stated for the plane slit along any ray. Theorem 1.8 is due to Loewner [147], where a more detailed study of univalent maps of slit regions appears. The following paper in the same journal (by Frank and Loewner) contains an application of these results to a hydrodynamical problem.

7. Bieberbach Conjecture

268

Another theorem on slit regions which may be proved similarly is" T h e o r e m 1 9 If h(z) = z + ~-~c~ Cnz_n is univalent in C - {z " - p < z < O,p > 0}, then Re C1 < O. 9

n--0

"

--

Proof. The map z --+ ~4 (~-~1)2 maps C - B(0,1) onto C - {z" - p < z < 0} and is univalent in C - B(0, 1). Hence 4h(P(z~_l)21 H(z) - p \ - . . . /

=

(z-l)2

z

=z+

4

+-C po

C0-2

4

z

+-C p 1 ( z - 1) 2 + . . .

1

+-+--C1(1+-+ z

pz

11 z

-~

+...

+... (1.13)

is univalent in C - B(0, 1), and so H E E. Clearly the terms involving C,, v _> 2 as coefficients do not contribute to the coefficient of ~1 in the Laurent series expansion of H around 0, and so we get from Theorem 1.3 and (1.13), 1 + ~C1 < 1 . Hence Re

(') -~C1

<_ O, or since 4/p is real, Re C1 <_ O.

D

Parallel to Theorem 1.7 we may also inquire about inequalities for arg f'(z). T h e o r e m 1.10. If f E S, then

1 § Izl) l argf'(z)l < 2 Log 1 Izl " Proof. If arguing from inequality (1.6) of the proof of Theorem 1.7, we use Jim wl <_ Iwl instead of IRe w] <_ Iwl, and replace z0 by ~ we are led to -4[ffl (r162 1 -ICI 2 < I m f,(ff)

< 41~1 - 1 -ICI 2 "

(1.14)

But, writing ~ = re i~

Im

( ~f'' f,(r(~) ) -

Im(r-~rd Log f, (rei~

d f, (rei~ r~rr(arg

and so (1.14) becomes -4 d 4 < 1 - r 2 < ~rr arg f " " i r i~ e 1 - r2 "

,

269

7.1. Elementary Area and Distortion Theorems

Integrating from 0 to Izl, gives - 2 Log

( 1 + zl) (1 + Izl) i - z I <-- argf'(z) _ 2 Log l -Izl " I-1

N o t e 1.7. Theorem 1.10 is a result of Bieberbach [25]. For the rotations of the Koebe function (lq_e~Oz)~, f'(z)

=

1 - e i~ (1 + e iOZ) 3

"

Also, since for z r 0 the points 1 + ei~ belong to the circle C(1, Izl), for z 7~ 0, I arg(1 +

e~~ <_ arcsin Izl,

and so for z r 0, arg f' (z) - arg(1 - e i~z) - 3 arg(1 + e/~z) _< 4 arcsin = 4 fo izl

Izl

1 1 dr < 4 folZl --------dr - 2 Log (1 + Izl' ~ (1 -r2) 89 1 -r 2 -1- - ~ ]

Thus (for z ~ 0), Theorem 1.10 is not sharp for the rotations of the Koebe function. In fact, it is not sharp at all. Sharp estimates for l arg f'(z)l were found by Golusin [86]. These are I argf'(z)l __ 4arcsin Izl if Izl _< ! argf'(z)l _< 7r + Log 1 - I z l 2 if

< lzl < 1 .

For further sharp bounds on arguments associated with univalent functions, see, for example, Chapter IV of Golusin, Geometric Theory of Functions of a Complex Variable [85]. It is of some interest to ask what results are obtained if suitable restrictions are made on the image domain of B(0, 1) under an f E S. For example, since the Koebe function does not map B(0, 1) onto a convex region, we might ask after Theorems 1.5 and 1.6, what happens if in Theorem 1.5, we required the image to be convex. T h e o r e m 1.11. Suppose f E S and the image T of B(0,1) under f is convex, then the disk B(O, 1/2) C_ T, and in general, this is the largest such disk centered at O.

7. Bieberbach Conjecture

270

Proof. The proof depends upon Theorem 1.5. Suppose f ( z ) r C for all z E B(0, 1). We claim that if g(z) -

(f (z) - c)

,

then g is univalent. For g is clearly analytic in B(0, 1) and if for some z and w in B(0, 1), g(z) - g(w), then f (z) - C - + ( f (w) - C), and so since f is univalent, if z r w, then f ( z ) 7~ f(w), whence we have f ( z ) + f(w) - 2C. But then since T is by hypothesis convex, l(z)+l(w) _ C E T; contradicting the definition of C since T is the image of B(0, 1) under f. If f ( z ) - z + a2z 2 + . . . , then g(z) - (f(z) - C ) 2 - C 2 - 2Cz + (-2Ca2 + 1)z 2 + - . .

and so h(z) - g(z) - C 2 -2C is univalent and h E S. Since by definition of C, g(z) 7~ 0 for any z E B(0, 1), 1 h(z) 7~ C/2 for any z E B(0, 1). Since h E S, by Theorem 1.5, [C/2[ >_ ~, and so [C[ ___ 1/2, which proves the theorem. The function ~ shows the result is sharp since it maps B(0, 1) onto the halfplane {z" Rez > - 1/2}. V1

As a parallel to Theorem 1.10, we have T h e o r e m 1.12. Suppose f e S and the image T of B(0, 1) under f is convex, then [ argf'(z)[ _< 2 arcsin [z[, and there is a function f E S with convex image T so that equality holds for a point z0 of B(0, 1). Proof. We use Theorem 1.5.5 (The Schwarz-Christoffel Formula for B(0, 1)). Suppose f maps B(0, 1) onto the Jordan interior of a convex polygon gln with n sides, then, by Theorem 1.5.5, f (z) - C1

n (w - Bk )~k-l dw + C2.

(1.15)

k=l

Since f(0) - 0 and f'(0) - 1, we get C2 - 0 and C1 - H2=~(-Bk) l-~k. Since the points Bk which go onto the vertices of the polygon all lie on C(0, 1), we can write Bk - e -iOk, Ok real. Finally, recalling that the quantities 7rak are the interior angles of the polygon gln, convexity requires ak _< 1, and the formula for the sum of angles of a polygon gives n

k=l

7.1. Elementary Area and Distortion Theorems

271

Taking these considerations into account (1.15) becomes

f(z) =

I I (1 - eiO~w)*"~d w ,

(1.16)

k=l

where #k -- ak -- 1 _< O, and s #k -- --2. Since I arg(1 -e~O~z)l <_ arcsin [z[ (all the points 1 -eiOkz lie on C(1, Iz[)), we have from (2) n

I argf'(z)l -

#k arg(1 -- eie~z) <_ 2arcsin Izl.

Z k=l

Since this last inequality does not involve the number of sides of the polygon, and since any convex set may be arbitrarily approximated by a polygonal convex set, it holds for all convex T. The function f (z) - z again shows that the result is sharp, as f'(z) = (1_~)2 1 and so arg f'(z) = - 2 arg(1 - z) which takes on the value 2 arcsin Iz] for suitably chosen z (which in fact can be explicitly computed). El N o t e 1.8. The elegant proof of Theorem 1.11 given is due to MacGregor [150], although the theorem had been long known. Theorem 1.12 is a result of Bieberbach in his already cited paper of 1919. Many results similar to those in Theorems 1.51.12 can be found by similar methods; for some further examples, in addition to the literature already cited, see Chapter V, Section 8 of Nehari, Con.formal Mapping [168].

()

( )4

T h e o r e m 1.13. If f E S, and Zl, z2 E B(0, r) where r < 1, then 1-r

4

l+r

/'(Zl)[ < f' < -

l+r 1

Proof. Since f is univalent in B(0, 1), f' does not vanish in B(O,r), and hence by the maximum and minimum modulus principles, and Theorem 1.7 for all z E B(O,r),r < 1, 1-r l+r (1 + r) 3 <- [f'(z)t _< (1 - r) a ' and so the result follows.

V1

N o t e 1.9. The function (l__Zz)= again shows the result is sharp. Koebe proved a more general distortion theorem of the same sort as Theorem 1.13, but with less precise bounding functions: If R is a region in the plane and D another region such that b C R, then there is a positive constant M, depending only on R and D, such that if f is univalent in R, then for any zl, z2 E D,

1

If'(zl)l
272

7. Bieberbach Conjecture

This result is of importance in proving Koebe's famous results on uniformization (which were established in 1907-1909 in a series of papers in Ghttingen Nachrichten; about which there is copious literature, but which cannot find place here. The above result is stated in [135]. N o t e 1.10. The theorems of this section dealing with limits on distortion of B(0, 1) under univalent maps f, or the size of disks in the image centered at f(0) are clearly different from those of Chapter 2, where we were examining the largest disks, necessarily contained in the image Z of B(0, 1) under an analytic or univalent map f, which were centered anywhere in/7. The difference between disks centered anywhere and disks centered at f(0) is considerable (cf. for example the image z of B(0, 1) under (1-z)2)" Nevertheless, as Landau first noted in 1922, there is a connection which can be used to establish distortion theorems for analytic maps which are not necessarily univalent. As an example of Note 1.10, we have the following result. T h e o r e m 1.14. If f is analytic in B(O, 1), f ( 0 ) - 0, f ' ( 0 ) - 1, then the image of B(0, 1) under f contains a disk B(O,p), where p is a positive constant. Proof. Suppose f is analytic in B(0, 1), f(0) - 0, f'(0) - 1, and f omits both the points e i~ and ae ir for some real 0, r and a ~- 1. Let n be an integer >_ 2 and let

(,,z, "-- (OZe i ( ~ 2 - 0 )

= ~l/ne~ 9

-- (e -iO

-- ozei(r

z

. . .+.

-F " " " ) l / n

--_

( 1 + - 1( e -i~ _le -ir )...) z+ n

.

Then, F is analytic in B(0, 1), and F does not take either 0 or 1 for z E B(0, 1). Hence, letting a - ctl/ne~(~-e) ~ , and b Theorem 2.2.3, for example, that

~/.~ e ~(~o-o) . (e-iO

1 e-ir

we have by

2Imv(a) 1 <_ Iv'(a)l Ibl '

(1.17)

where v is a fixed branch of the inverse of the elliptic modular function #. Taking for simplicity a = 2 -n and r = 0, and setting L - 2 lm,(~) _ 2 sin,( 89 we get from (1.17),

,

1 L :> I b ] - ~nn(2" -

iv'(a)l

-

Iv'( 89

'

1),

and thus if n is so large that 2~____A1> L we have a contradiction. Hence, we get that if n is the least positive integer such that 2"-1 > L, then the image of B(0, 1) under f contains the disk B(0, 2-~). [--I

7.2. Some Coefficient Theorems

273

N o t e 1.11. The above result is due to Landau [138]. For yet another covering theorem of the same sort, under the auxiliary condition that for z e B(0, 1), f(z) ~ 0 for z ~ 0, see Theorem 8.6.11. Actually, many results on functions univalent in B(0, 1) can be made to yield results on functions analytic in B(0, 1), by what is known as the principle of subordination introduced by Littlewood in 1925. This principle is dealt with in almost any discussion or book on univalent functions. See, for example, Chapter VIII, section 8 of Golusin's cited book, or Chapter 2, section 2.1 of Pommerenke's. Both of these contain references to further literature. There are also generalizations of distortion theorems to multiply-connected regions, and to functions of several complex variables.

7.2

Some

Coefficient

Theorems

Let us note first T h e o r e m 2.1. The family S is compact; that is, if {fn) is any sequence of ]unctions in S, then { s contains a subsequence converging for all z e B(O, 1) to a

]unction of S. Proof. By Theorem 1.7,

Izl

[f(z)l ~ ( 1 - Izl)2 ,z e B(O, 1). Hence the family S is locally uniformly bounded for all z E B(0, 1), and so by Theorem 1.4.2, S is a normal family, and {fn} has a convergent subsequence. By Theorem 1.2.6, a convergent sequence of univalent functions converges to either a univalent function or a constant. Since for all f E S, f~(0) = 1, the limit function cannot be a constant, so it must be a function f univalent in B(0, 1), and, as is easily seen, satisfying f(0) = 0, f'(0) = 1. [-1 N o t e 2.1. It is worth noting that since B(0, 1) can be carried onto any other disk by a MSbius transformation, and since a family of functions is normal in a region 7~ if and only if it is normal in all disks C_ T~, we get by the above argument that any family of functions univalent in a region T~ is a normal family. Furthermore, given a point ~ E P~, the subfamily of this family consisting of all functions such that If'(~)l _ C > 0, is compact by the above argument. In fact, any condition ruling out constant limits will suffice to demonstrate compactness. As an immediate consequence of Theorem 2.1, we have T h e o r e m 2.2. There exists a ]unction F,~ E S, such that An - supfes ]nth coal.ficient of fl - Inth coefficient of F,~I.

274

7. Bieberbach Conjecture

Proof. Let J ( f ) - [nth coefficient of f[. Clearly there is a sequence of functions fk,n E S such that l i m k ~ J(fk,n) - A,~. Since S is a compact family, there is a subsequence {fk~,n} of {fk,n} converging to a function Fn E S, and, as is easily verified, J(F,~) - v---+(:x:) lim J ( f k , , n ) - A n .

[3 After the results of Section 1, in which the function ~ frequently played the role of extremal function, it is reasonable to conjecture that it might well be extremal for the problem of the maximum modulus of the nth coefficient (as it is for la21). This is the Bieberbach conjecture. Precisely Definition 2.1. (The Bieberbach Conjecture) is that if f (z) - z + ~"~n~=2anz n e S, then [anl <_ n. The Bieberbach conjecture stimulated much research on univalent functions throughout the 20th century and was attacked by a great variety of methods since Bieberbach first posed it in 1916. In 1984 a proof was announced by Louis de Branges. The proof given here basically follows the version of his proof given by Fitzgerald and Pommerenke in 1985 (see also Conway [51], vol. 2). A basic tool of this proof goes back to 1923 when it was proved by Karl LSwner (who later immigrated to the U.S. and Anglicized his name to Charles Loewner). LSwner used his approach to prove the Bieberbach conjecture for a3. We begin thus with the definition of Loewner chains. Definition 2.2. A Loewner chain is a continuous ]unction f

9B(O,1) x [0,~) --+ C

such that

(i) for all t E [0, c~), f ( z , t) is analytic and univalent (ii) f(0, t) - 0 and ~ ( 0 , t) - e t

(iii)

if 0 < s < t < c~, f(B(O,

1),s) C_ f(B(O, 1),t)

etz

E x a m p l e 2.1. f ( z , t ) - (1-z)~ is a Loewner chain. Note that f(B(O, 1),t) is the slit region C-{z'z

Et

real, - c c _< z <_ - ~ - }

(see Example 1.1). N o t e 2.2. For any function g E S, f ( z , t) = etg(z) automatically satisfies properties (i) and (ii) of Definition 2.2, but property (iii) is not automatically satisfied. This is corrected by the following result which depends on the Riemann Mapping Theorem (Theorem 1.5.1).

275

7.2. Some Coefficient Theorems

T h e o r e m 2.3. Suppose {f~(t) 90_< t < c~} is a family of nested simply-connected regions (i.e. for 0 < s < t < oo, f~(s) C f~(t)) such that as n ~ c~, f~(tn) --+ C. For each t > O, let ht be the inverse of a Riemann mapping between B(O, 1) and f~(t). So ht " B(0, 1) -~ f~(t) and we can prescribe ht(O) - O, h~(O) - B(t) > O. Let ht(z) - h(z, t) and ~(0) - ~o, then (a) fl is a continuous strictly increasing function and/~(t) ~ co as t --+ co. (b) Let A(t) = log ~

)

and f ( z , t ) -

Loewner chain with f(B(O, 1),t)

-

~1 h(z,,~-i (t)); then f defines a

~Ft(~-t(t)).

Proof. Note that if tn --+ t, then ht. --+ ht and so fl(tn) -+ ~(t); hence fl is continuous. Now if s is fixed and < t and ft(s) C Ft(t), then there is an analytic function taking B(0, 1) onto itself such that hs(z) = ht(r for all z E B(0, 1), and r = 0. Then Ir < Izl by Schwarz' Lemma, and in fact, because f~(s) ~ f~(t), Ir < Izl. Also, here Ir < 1 (again because of the inequality). Hence ~(s) - h'(0, s) - h'(0, t)~b' (0) - ~(t)~' (0). Now i~' (0)i < 1 and so for s < t, B(s) < /~(t) so ~ is a strictly increasing function from [0, oc) to [0, co). Furthermore, since f~(t) -~ C as t ~ co, ~(t) ~ cc as t --+ oc. So /~ 9[0, co) ~ [B0, co) is a continuous strictly increasing, and onto map.

So A(t) - l o g

(~-~o t ) is a strictly increasing function taking [0, c c ) o n t o itself.

Then, f ( z , t ) - ~ h ( z , A - l ( t ) ) i s a continuous function which is clearly analytic and univalent for all t > 0. Also, clearly f(B(O, 1),s) C_ f(B(O, 1),t). Clearly f(O,t) - O, (ht(O) 0). Also, if u - A-l(t), then t - A(u) and so e t = -

B(u) ~o

~o

-

so ~ ( f ( z t))

-

et

z=O

(since Oh h(z t)

(t)). Thus f is a Loewner chain.

- ~(t)) z=O

Clearly f(B(O, 1) t) [-1

N o t e 2.3. Clearly ~ must appear in the result since for an arbitrary region gt, there is no reason to think that the map from B(0, 1) to f~ (inverse of the Riemann map from f~ to B(0, 1)) is in class S, while for a Loewner chain f ( z , 0) is in class S. An important example is' E x a m p l e 2.2. Suppose 3' is a Jordan arc that does not pass through 0 and such that 3,(t) -~ oc as t --+ co. Let 7(0) - a0. For 0 < t < c~, consider 7t, the restriction of 3' to [t, oc) and let f~(t) - C - ~t. Then, by Theorem 2.3, we can find a Loewner chain f ( z , t). N o t e 2.4. A Loewner chain can be thought of as a parametrized family of univalent functions ft(z) - f ( z , t ) starting at f0 and as t --+ co, the ranges of the functions expand to fill the whole plane. T h e o r e m 2.4. For every ]unction fo E S, there is a Loewner chain such that f ( z , O) = fo(z) in B(O, 1).

7. Bieberbach Conjecture

276

Proof. Assume first that f ( z , t ) is analytic in a neighborhood of B(0, 1). then 7 - f(C(O, 1)) is a closed Jordan curve. Let g ' C o ~ - B(0, 1) --4 C ~ - (closure of the Jordan exterior of 3') where g(oc) - cc and g'(c~) > 0. Let, for 0 <_ t < co, f~(t) - the Jordan interior of the curve g({z "1 z] - et}), so f~(0) - fo(B(O, 1)). Also, clearly for 0 <_ s < t < oc, ft(s) C f~(t). Then by Theorem 2.3, in the notation there, let h(z, O) - fo(z) and then ~0 - 1 (this is true by the uniqueness of the mapping function). Theorem 2.3 then proves Theorem 2.4 in this case. In general, for an arbitrary function f E S, put for each positive integer n, rn - 1 - !n and suppose fn(z) - !7 ' nj ' ( r n ~ z) (so In(z) E S) " Then clearly each j~n is analytic in a neighborhood of B(0, 1). So there is a Loewner chain Fn with Fn (z, 0) - f~ (z). It is easy to prove that the set of all Loewner chains is compact. So some subsequence of {F,~} converges to a Loewner chain F and then F(z, O) - f ( z ) in B(0, 1). [-I N o t e 2.5. Suppose fn(z) - f ( z , 0) maps B(0, 1) onto the complement of a Jordan arc extending to c~, then Theorem 2.4 is just Example 2.2. D e f i n i t i o n 2.3. s is the set of all Loewner Chains. T h e o r e m 2.5. If f E s and 0 <_ s <_ t < oe, then there is a unique analytic ]unction r s, t) defined for z E B(0, 1) such that (i) r (ii) r ~162

s, t) E B(0, 1) and f ( z , s) - f(r

s, t) is univalent, lr

s, t) - 0, r t))

s, t), t) for all z E B(0, 1) s, t)] <_ Izl (for all z E B(0, 1)),

z--0

(iii) For all z E B(0, 1), r (iv) If s <_ t <_ u, then r

r s, u) = r162

s, t), t, u) for all z E B(O, 1).

Proof. Since f ( z , s)(B(O, 1)) C_ f ( z , t)((B(O, 1)), there is a unique analytic function defined on B(0, 1) such that (writing fu(z) - f ( z , u ) ) , ft(r s, t)) - fs(Z) and r s, t) - 0. Since ft and fs are both univalent on B(0, 1), so is r By Schwarz' Lemma ir s, t)l _< Izl for all z E B(0, 1). Taking the partial derivative with respect to z at both sides of (i) and evaluating at 0 (denoting the derivative by ')

S'(0,

- S'(r

t), t)r (0,

t).

But f'(O,s) - e ~ and I' (r s, t), t) - e t so r - e s-t. Since r is unique, r s, s) must - z for all z E B(O, 1) (since f ( z , s) - f(r s, t), t)).

7.2. Some Coe]ficient Theorems

277

Again, the uniqueness of r shows that if w - r f(r

t, u), u) -

s, t), then

t) =

t)

and so (iv) follows.

W1

D e f i n i t i o n 2.4. The function r

defined for z E B(O, 1) of the preceding theorem is called the transition ]unction for the Loewner chain.

N o t e 2.6. The transition function r in fact, establishes its uniqueness).

s, t) is given by r

s, t) - f t l ( f s (z)) (this,

From now on, we consider only the situation in Example 2.2 (see also Note 2.4). So, to restate: 7 " [0, co) --+ C is a Jordan arc with 7(0) = ao that does not pass through 0 and 7(t) ~ co as t --+ co. For 0 _< t < co, ")'t is the restriction of 7 to It, co) and ~(t) - C - 7t. We assume there is a Loewner chain f such that ft(B(O, 1)) = ~(t) for all t ___ 0 and ~o - 1 (see Theorem 2.3 and Note 2.2 above). Let gt - f[-1. ~(t) --+ B(0, 1) (so in the notation of Chapter 1, gt is a Riemann mapping function). Define g((, t) = gt((). Let r be the transition function for the chain f and for s < t, - r s, t). Recall that Cst is unique and - ft--1 o f~. The functions j's and .ft have continuous extensions to C(0, 1) (this, in fact, is an application of the Osgood-Caratheodory Theorem mentioned Chapter 1; but which is not proved in this book). Similarly gt has a continuous extension to ~t(t) t2 {-~(t)}. There is a unique point on the unit circle, say ~(t) such that .ft()~(t)) - 7(t). Let "),([s,t]) be % ")'t and let Cst be the closed arc of C(0,1) defined by {z" fs (z) e ~([s, t]) }. Let Jst - gt (7Is, t]). So Jst C_ B(O, 1) U )~(t) and is a Jordan arc. So Cst maps B(0, 1) onto B(0, 1) Jst, Cst has a continuous extension to B(0, 1) which maps Cst onto Jst and C(0, 1 ) Cst onto C(0, 1) - {~(t)}. Note that )~(s) is an interior point of Cst and as t --+ s, Cst shrinks to ~(s) while for fixed t and s --+ t, J~t shrinks to A(t). Let Js*t b e the reflection of Jst across the unit circle, then by the Schwarz Reflection Principle (see Appendix) C - Cst can be conformally mapped onto C (Jst U Js*t); this is an analytic continuation of Cst to C - Cst, which will continue to be denoted by est. With the preceding notation, we have (a) F o r z E C - C s t ,

T h e o r e m 2.6.

_< 4e t-s.

!r I

--

!

( b ) ) ~ " [0, co) --+ C(0,1) is continuous.

Proof. Jst C_ C - Cst(C(O, 1)). Now, lim r 9

z

s, t) = lim

r

z _ s,t) r

1

= et_ s

278

7. Bieberbach Conjecture

(by Theorem 2.5). So, by Theorem 1.5 (or 6.5.4), every point in Jet is contained in the complement of {z ]z[ 9 < lr s,t)}, i.e. has a distance of at least ~~r s , t) eS-t

from boundary of r 1)). So J,, C_ C - { z ' l z ] < -W-}" So for Js*t, which is the reflection of Jst across C(0, 1), we have

Y~,t c_ {z. [z[ _< 4et - ' } . This proves (a). 1 For (b), note that (a) shows that for any T >_ s, {~r 9s _< t _< T} is a normal family (Section 1.4, particularly Theorem 1.4.2). Hence, by the uniform boundedness, every sequence has a convergent subsequence to an analytic function. So if {r } converges to say r analytic on C - {A(s)} as tk -4 s (t k > S). Then r is bounded on C - {A(s)}, so A(s) is a removable singularity, so r must be constant. Furthermore, r - limt_~s r s,t) - 1. So r z __4 1 uniformly on compact sets of C - {A(s)} as t --4 s(t > s). So r -~ z uniformly on compact sets of c - {~(~)} as t -~ ~(t > ~). Fix s _> O. Given e > O, choose 5 > 0 so that for s < t < s + 5, C~t C_ B(A(s), e). For C - C(A(s),e), let X - Cst(C), so that the interior of X (a Jordan curve) contains both Jst and Js*t and hence also contains A(t). Now for sufficiently small 5, [ r z[ < e for all z E C. So for z E C, IA(s) - A(t)l

< I:~(~) - zl + Iz - r

+ Ir

- ~(t)l < ~ + ~ + Ir

- A(t)l.

But [r - A(t)[ is less than the diameter of X, and so less than 3e (for all t sufficiently near s). So, [A(s) - A(t)] < 5e. So A is right continuous. The proof that A is left continuous is exactly similar. This proves (b). Now r for z E B(0, 1) is zero only for a simple zero at 0 so ~ has no zeros in B(0, 1) and so we can define the analytic function

o/z by choosing the branch of the logarithm where ~(0) - s - t. (Recall that r (0) = es-t). ~(z) is analytic on B(0,1) and continuous on B(0,1). Now Cst(Cst) = the Jordan arc Jst and if z E C(0, 1 ) - Cst, then r E C(0, 1 ) - {A(t)}. (See i

discussion before Theorem 2.6); so R e r

__

!

- log ] r (~) ] - 0, everywhere i

on

C(0, 1)

!

except Cst. Now, if we let e i~ and e i~ be the endpoints of Cst, then, by the Poisson formula, R~ ~(z) -

Re ,~(ei~

e i~ + z ) e iO - z dO.

But having chosen the branch of the logarithm where ~(0) = s - t, so Re ~(z) - O, we get

r

e i~ + - ~1 L ~ R~ ~(~Ole~o : zz dO.

(2.1)

7.2. Some Coefficient Theorems

279

Note that then

9(0)

-

~1

f~

Re r

(eiO)dO -

s-

t.

Now r - r s, t) - f t l ( f , ( z ) ) . (Note 2.5); or ft o r - f,. Since we defined gt - f[-1 and g({, t) - gt(r we get gt - r o g,, and putting z - g,(~), get that log(Cst(z) ) - l o g gt(()

z

;

so, by (2.1) loggt(~) _

(~----~ gs

1 /a ~

Re ~(e i~

e ie + gs(~) ) dO ei o _ gs(() .

We break this last integral into real and imaginary parts of the integrand and apply the Mean Value Theorem to each part. Since, as already observed,

2-7

Re(~b(eiO))dO - s - t

this gives log

gt(r

_ (s _ t)

Re

-.

+ i lm

where a _< u _ s); then e iu and e iv both converge to ,~(s); so lim 1 log (gt(~) ~ _ ~(s) + gs(~) t-~s t- s \ - A ( s ) - gs(~) " t>s Let x(s) - X(s). Then, since A(s) E C(0, 1), we have lim.1 log(gt(~'))_ l+x(s)gs(r t-+, t- s g,(r - 1 - x(s)g,(~) " t>s But the left-hand side of this equation is the derivative (from the right) of loggt(s) at t - s. A similar argument shows that the same formula holds for the derivative from the left. So we get 0

0-7(g,(O)

t--S

_ _g~(~) ( l + x(s)gs(r ) 1 - x(s)gs(~) '

or since gt(r - g(~, t),

t--s

- -g(~, s) ( l + x(s)g(~' s) ) 1 - x(s)g(r s) "

This last allows a proof of Loewner's differential equation.

(2.2) 1"1

7. Bieberbach Conjecture

280

T h e o r e m 2.7. Let ] be a Loewner chain such that fo is a mapping onto a "slit

region" (as in Example 2.2). (Such a mapping will be called a "slit mapping".) Then there is a continuous function x : [ 0 , c~] -~ C(0,1) such that f(z, t) exists and satisfies Of ( l + zx(t) ) 0 -~ (z,t) - 1 - zx(t) Z-~zf(Z't) " Proof. Let F(z, t) from C(0, 1) x [0, c~] to C x lI~ be defined by F(z, t) - (f(z, t), t). The image of F is the open set

[.j (c ~) -

x

(t, ~)

t>o

and F is a one-to-one mapping whose inverse is F - I ( ~ , t) = g ( ( ~ , t ) , t ) . So F -1 is continuously differentiable with Jacobian

-~g(~,t)

O11'

which equals ~ g ( ~ , t) which we know ~t 0 (in fact

=

1 ~I(g((,t),t) ) " It follows that

F is continuously differentiable, and so ~ ( f ( z , t ) ) exists and is continuous. Let x be as above. Since = f(gt(~, t)), differentiation gives

0With z

=

a 0 N](g~(~l,tlN(g~(~))+

0 Nl(g~(r

gt(~) - g((, t) (cf. argument before Theorem 2.7); this becomes

_cOt f (z, t ) - _ -~zf (Z, t ) ~ g ( ,

t) -

0 1 + x(t)g(~, t) _ =O'-~](z't)g(~'t) l - x(t)g(r -

(2.3)

1+~(t) o f

=z 1

-

zx(t) Oz (z,t) .

Equation (2.3) is called Loewner's differential equation.

[-7

It is now necessary to consider relations on the coefficients of power series which are exponentials of power series and their relations to the original coefficients. The idea of utilizing this relationship in some manner is quite old as is the use of

281

7.2. Some Coefficient Theorems

Loewner's equation (which Loewner himself used to prove ]a3[ _< 3 in Bieberbach's conjecture). Suppose r is an analytic function in some neighborhood of 0 with r - 0, and oo

~(z) - ~

~z *

k=l

its power series. Let r - e r - ~~ o bkz k. With this notation we claim that Theorem 2.8. _ (n + 1) exp ( ~ ' 1 n ]bk 12 < ~ ( n +Ll k-0

klakl 2 - ~ ~-)) 1 .

m---1 k = l

Proof. oo

kbkz k-~ - ~ r 1 6 2

r162

k--I

-

(k+l)ak+~

Zk

k=O c~

: ~

~bkz

k

-

k=O k

}2 (m + lla~§

k=O m=O k

So (k + 1)b~+l - ~ ( m

+ llam+lbk-m.

rn--O

Replacing k by k - 1 and m by m - 1, this gives k

kbk - E

mambk-m .

m=l

So, by Cauchy-Schwarz, k

k

k

k-1

elb~I~ <- 5: m~lo~i~ ~ Ib~-~i~- ~ m~io~i~ ~ Ib~i~. m--I

m----I

m--I

m--O

(2.4)

282

7. Bieberbach Conjecture

Now by (2.4), Ib~l~ = ~ b~, + b~ _< k=0

k=0

~-i

B

(

Ibk 12 1 +

1 n -'~ E

k=O

m2 tam

12

m

n

m=l

n-l~ 1( __lb~l ~n+ i n k=0

1 + n+l

.(. + 1)

m21aml 2

)

.

m=l

But exp(x) _> 1 + x, so !2

Ibk <_ k=O

(~

Ibkl

Y~"n=lm=lainI= - n

2) ~n +e xl p n

)

n(n + 1)

\k=O

(2.5)

n-1 Repeating for ~k=0 b2 gives

n-1

k=O

~k=0

)

n - 1 exp

(

n ( n - 1)

(n- 1)) .

(2.6)

Combining (2.5) and (2.6) gives n-2

k=O

Ibkl2 <_ n + 1 n- I ~

k=O

( n-1 12 Em=l m21aml 2 - ( n - 1) + Ibk exp n ( n - 11

n

Em=l

2

2

m a m -- n

n(n + 1)

,

Continuing in this way, gives n

k=0

Ibkl2 <_ (n + 1)exp B Zk=l m2tarnl2 - k k'-I k(k + 1) -- (n + 11 exp

(t 1 s (t 1 k(k + 1)

k=l

-- (n + I)exp

=

k(k + I)

k----1

--

m21amt2- B

1)

(2.7)

k+i

k=l

m~l~l ~ + I- ~ ~ k--1

.

k 1 _ But, writing Ak for ~m=1 m21am [2 , by partial summation {since ~'~'n~k=1k(k+1) --

7.2. Some Coefficient Theorems

283

1 1 - h-gY) we get

k=l

k(

A~

1) =

~(

1 )(Ak_Ak+l)+( 1

1

k=l

= E-((k k=l n

k2t_ 1

+ 1)21ak+l

)

1

1)(

k+ 1 +

n-I k=l

n

n

1 ) " j2 ~--'k2lak k=l 12

n+l

n

n + 1 E k2Jak k--1

.1l

= ~ k21akl2 - ~ k(k - 1)la~l ~ k=l

1

1

= ~ k21akJ2 - ~ k(k + 1)lak+,[ 2 k=l

1)Ao

n+l

n + 1 E k21akl=

k=2

k=l

n

n

= ~ klakl2 k=l

1 12 n + 1 E k2la~ " k=l

Thus in (2.7), the quantity being exponentiated is equal to n

Eklakl 2 k=l

n

n

1 E +I k=l

1

n+l

k~lakl = + 1 -

E

k=l

Writing this as a fraction with denominator n + 1, it equals 1 n+l

l

n+l 1

(~

(n + 1)k]akl 2 - k

k=l

k=l

~ { (n + l)klak[2 _ k2[akl2 + 1 n

1

k--1

E

E

m=l

k=l

12(n + l

-

n+l}_ k=l

k=l

n + 1 E klak n+l

k2]akl2 + n + 1 - ~

) +n 1 k

--

k

k)_n+l_k k

klak]2 - - ~ 1

) 9

Combining this with (2.7) proves Theorem 2.8.

[3

Note 2.7. All the above was known many years before 1984--the Loewner differential equation since 1923. Milin and Lebedev investigated the relationship between the coefficients of a series and those of its exponential in the 1960's and versions of the Milin-Lebedev inequalities (though not Theorem 2.8) appear already in Pommerenke's 1975 book on Univalent Functions [203]. Theorem 2.8 appears as the "Second Lebedev-Milin Inequality" in Duren, Univalent Functions [61]. It appears to have been first announced by Milin in 1967 but a proof waited until his book

7. Bieberbach Conjecture

284

of 1971 (that was not translated into English until 1977). There are two other "Milin-Lebedev" inequalities, but for DeBranges' Theorem only the second one is necessary. Readers interested in these should consult either Duren's aforementioned book or Volume II of Conway's Functions of a Complex Variable [51]. Milin had made the conjecture that if f E S and h is the branch of l log ( / - ~ ) with h ( 0 ) - 0; and oo

h(z) - ~ ~ z ~ n=l

n m(

then for all n >_ 2,

E

E

m=l

k=l

kl"/k[2 - ~ 1

)

_
Milin's conjecture, if true, implies Robertson's conjecture that if g E S and is odd and has the power series expansion oo

g(z)

z+

--

c2k+1z 2 k + l

E k--1

then

n-1

1+ ~

Ic2~+ll 2 < n,

k--1

and this in turn implies the Bieberbach conjecture. Fitzgerald and Pommerenke have shown that DeBranges' ideas cannot be applied directly to a proof of the Bieberbach conjecture. To make use of Loewner's differential equation, let f be a slit mapping in S and F the Loewner chain with Fo - f, e-tFt E S for all t >_ 0. Define h(z, t ) -

-~ log

et z

- E 7k(t)zk , where h(O, t ) = O. k=l

Define the function r

E

klTk(t)[2-

Tk(t)

k=l

where the Tk (t) are certain functions introduced by DeBranges. They will be defined later, but the properties that they have that will be used are:

(~)

~k(t) - ~k+~ (t) -

(b) T k ( 0 ) = n + l - - k (c) limt~oo Tk(t) -- 0 (d)

~-~(t)<0

- ~ , ~k( ~ )

-

~+~(~) k+l

7.2. Some Coefficient Theorems

(e)

285

Trtq-1 ( t ) - - 0

In other words, DeBranges applied "weights" T(t)to the Milin coefficients,and the heart of his proof is to show -~t > 0. As above, we shall use ~ to indicate differentiation with respect to t and 0 to indicate differentiation with respect to z even when there is only one variable in the argument. Of course, in Loewner's differentialequation, there are two variables to be concerned with.

Theorem

2.9.

r

_> 0.

Proof. By Loewner's differential equation (Theorem 2.7) and the definition of 7~(t), E-~ k=l

"/k(t)zk= 1 0 log 20t

F(z,t) etz

1 (OF(z,t))1 = -2 -F-(~,, t ) - 1

But, we know that

Ix(t)!

(2.8)

(l+zx(t)OF(z,t)) - -~ z l _ z x ( t ) -~z,, t ) - 1

.

= 1 and so for Izl < 1, (2.9)

1 + x(t)z 2x(t)z = 1 + 2 E ( x ( t ) ) k z k . 1 - x(t)z = 1 + 1 - x(t)z k=l

On the other hand, with h(z, t) - ~ k = l 7k(t) zk

1 (~ Z

So

~F(z,t) F(z,t)

-

kTk(t)z ~-1 k=l

1 o0 = - + 2 E kTk(t)zk-1 z k=l

Substituting this in (2.8) and using (2.9) gives

F_,

-

-~ z

1 +2

(x(t))kz k

- +2 Z

k--1

kTk(t)z k-1

- 1

.

k--1

Thus we get 1+ 2E k----1

-~ ~/k(t)zk -

1 + 2 E(x(t))kzk

1 -t- 2 E

k---1

kTk(t)zk

k--1

Comparing coefficients in (2.10) gives k-1

-~Tk(t) - kTk(t)+ x(t) k + 2 E rx(t)k-~Tr(t) " r=l

"

(2.10)

286

7. Bieberbach Conjecture

Let b k ( t ) - E k~=~ r x ( t ) - ~ % ( t ) for k >_ 1 and bo - O . Then, the above becomes (9

-~')'k(t) -- x(t) k - k')'k(t) + 2 E

k

r(x(t))k-r')'r(t) -- x(t)k - k~/k(t) + 2 x ( t ) k b k ( t ) ,

r--1

and we observe that (2.11)

bk(t) -- bk-1 (t) -- kx(t)-k~/k(t) . So

0 - ~ / k ( t ) -- x(t)k(1 -- bk(t) + bk-l(t) + 2bk(t)) - x(t)k(1 + bk(t) + bk-l(t)) . Now, clearly dklTk(t)12 -- k d~/k(t)~k(t) - 2 k R e ~ ( T k ( t ) ) z / k ( t ) dt

.

Now, since x(t) is on the unit circle, (x(t)) -1 - x(t), and so by (2.11) bk(t) - bk-1 (t) - kx(t)k~/k(t----~ .

Using this we get dt I~'k(t)

,2

= 2kRe

(0) -~,k(t)

7k(t)

= 2 k R e ( x ( t ) ) k ( 1 + bk(t) + bk-1 (t))(bk(t) - bk-1 (t)))

= 2Re(1 + bk(t) +

bk-1

1 k(x(t)) k

(t))(bk(t) - bk-1 (t)).

So, for the function r defined preceding the statement of Theorem 2.9, we have - ~

--Otr

klT~(t)l 2 _ kl ~rk(t)0

k----1

(2.12)

n

+ E 2 Re((1 + bk(t) +

bk-l(t))(bk(t)

- bk-1

(t)))Tk(t).

k--1

With m

Am - E 2Re(1 + bk(t) + bk-1 (t))(bk(t) - bk-1 (t)) k=-I m -

2Re E k=l

m

bk(t) - bk-1 (t) + 2Re E ( b k ( t ) k=l

- bk-1 (t))(bk(t) - bk-1 (t)),

7.2. Some Coefficient Theorems

287

we get since bo - 0, m

Am - 2Rebm + 2 ~

Ibk]2 -[b}_l[ 2 - 2Rebm + 2lbml 2 .

k=l

By partial summation, then, the second sum in (2.12) equals n

n

Ak(Tk -

T k - 1 ) -- 2 E ( R e b k

k=l

(2.13)

-t-Ibk[2)(Tk -- T k _ l )

k=l

(since Tn+l (t) -- 0 by (e), and where we have suppressed the dependence on t). For the first sum in (2.12) we have

n(O~-~(t) )( kI-y~(t)-,2

k=l

1(o )

= E-s

~rk(t)

k=l

~

-~Tk(t)

([bk - b k - l l ~ -1),

k=l

since Ix(t)]- 1. Thus, (2.12) now becomes r

1(o )

(k21~/k(t)l 2 - 1 ) - E

1(0 ) -~ -~T~(t)

- 2E(Rebk+[bkl)2(rk-Tk+l)+E

(Ibk-bk_ll2-1)

(2.14)

k=l

k=l

For the terms in the first sum here, we use (a) and so get

(Re bk + Ibk[2)(Tk(t) -- Tk+l (t)) -- - ( R e b k + ]bkl) 2 ( ~Tk(t)k

+

k+

Summing from 1 to n gives n

n

Re(bk + Ibkl2)(rk(t) - rk+~) - - ~ k--1

k---1

0

k (~t ) ~~ (r R

k

bk + Ibkl~ + R~bk_, + Ibm-, I~),

since bo - 0 and Tn+l (t) -- O. So (2.14) now is 0_0

ot r

n

- - ~

0

~-~(t)k (2Rebk + 21bkl2 + 2Rebk_l + 2[bk_~[ 2 - Ibk -bk-112 + 1)

k--1 n

k--1

0

k

(2Rebk + ]bk]2 + 2Rebk_l + [bk-~l 2 + 2Rebkbk'~ + 1),

288

7. Bieberbach Conjecture u

on writing [bk- bk-l[ 2 - (bk - bk-1)(bk - bk-1). It is easy to see in the same way that this last gives 0 -ot - r

n

OT -""'ff--[bk 4-bk-1

= - Z

q- 1

k--1

Yl

Since by (d) O r k ( t ) < 0, this proves Theorem 2.9.

To prove Milin's conjecture for functions in S that map B(0, 1) onto a slit region, note that by (b) Tk(O) - - n + 1 - k and so r

(

klTkl2 -

k=l

- E m=l

E k=l

kl%12

1 -- ~

)

"

Since by (c), Tk(t) --+ 0 as t ~ co, and by Theorem 2.9 ~ 1 6 2 > 0, we get - ; m--1

k--1

)

- r

-

/o

Nr

dt<_O,

which proves Milin's conjecture for a slit mapping f E S (recall that the 7n are defined by h(z) - ~]~=1 7~ tn where h(z) is the branch of l log ~ with h(0) - 0). Two issues for Milin's conjecture remain: (i) what about other f E S?, and (ii) what are the mysterious functions T~(t)? There is also the problem of how to deduce Bieberbach's conjecture from Milin's. Question (i) is easy to answer since the slit functions turn out to be dense in S. That is, precisely, T h e o r e m 2.10. To each ]unction f E S, there is a sequence of slit mappings fn e S so that fn -+ f uniformly on each compact subset of B(O, 1). Proof. Let f be a function in S. All that is necessary is to produce a slit mapping 9 E S so that

If(z)- g(z)l < e for lzl <_ p < 1 where e and p < 1 are given positive numbers. The theorem will follow by choosing sequences {r and {p,~} where cn --+ 0 and pn ~ 1. Suppose f E S maps B(0, 1) onto a domain D bounded by an analytic Jordan curve F. Let Fn be a Jordan arc connecting ~ to a point w0 E F then going part of the way around F to some point wn. If Dn is the complement of Fn, let gn map B(0, 1) conformally onto Dn with gn(O) - 0 and g~(0) > 0. Suppose the points wn are chosen so that Fn C F,~+I and wn ~ w0. Then clearly D is the kernel of the sequence of regions {Dn}, that is D is the largest domain so that 0 E D and every compact subset of D lies in all but a finite number of the domains Dn. Clearly every subsequence of {Dn} has the same kernel. Then, by the Carath6odory convergence

7.2. Some Coefficient Theorems

289

theorem (see Appendix) g, --+ f uniformly on compact subsets of B(0,1). But then g~ --->f' and in particular, g~(0) --+ f'(0) - 1. So ' g"(~) g" (o) E S and they are slit mappings converging to f uniformly on compact subsets of B(0, 1). This proves Theorem 2.10, and justifies the fact that Milin's conjecture need only be proved for slit mappings. !:3 There remains the question of the mysterious functions Tk(t) satisfying (a)-(e) (see discussion preceding Theorem 2.9). DeBranges' functions are defined (for n >_ 1 and l_
(*)

v--O

and Tn+l(t) -- 0; where, as usual, (z), - z ( z + 1)...(z + v - 1)

for any complex number and any non-negative integer v. This seems to make them even more mysterious, until it is realized that can be expressed in terms of the well-known J acobi polynomials. That is T h e o r e m 2.11.

_ffiTk(t)O

n-k

O~rk(t) - _ k e - k t E

p(2k,0)(1_ 2e -t)

v:O

where P(a'~)(x) is a Jacobi polynomial. Proof. Here the Jacobi polynomials are defined as usual as the orthogonal polynomials on [- 1, 1] with weight function (1 - x) a (1 + x) ~, a > - 1, fl > - 1. Here, as usual P(~'~)(1)-(n+a)-(n+a)(n+a-1)"'(a+l) rt

Tt!

"

Explicitly the Jacobi Polynomial ( x - 1)~-m(x + 1) m .

P(~'~)(x) - 2 -n E m=O

(2.15)

m

Askey and Gasper have proved that m

m (2k + 1 ) 2 . ( 2 k + 2v + 2)m_~

E p(2k,O)(x) -- E v--O

22~(2k + l),v](m - v)'

v=O

(2(x - 1)1"

(2.16)

"

(see American Jnl. of Math., 1976, pp. 709-737; p.717, where we have used the fact that (2k;1) (k+l)(2k+l)2v v

v --

22 v

7. Bieberbach Conjecture

290

for a = 2k in their formula.) The reader interested in the Jacobi polynomials is referred to the classic text by Gabor Szeg5 on Orthogonal Polynomials, published by the American Mathematical Society[226]. Now, n-k

ektk-~rk(t)O _ E (-1)"(2k + uul(n + 1)~(2k_ k - +u)!2u+ 2)n-k-~ e-~t

(2.17)

v=0

But, by (2.16) we have n-k

n-k

E P(2k'~

- E (2k ~ 1)2~(2k + 2u + 2)~-k-~

~=o

~=o 22 2k + 1)~u!(n - k-- ~)i (2(x - 1))

SO

n-s

p(2k'~

-

n-k

v=O

But

(2k + 1)2~(2k + 2v + 2)n-k-~ 22~(2k + 1 ) . v I ( n - k - u)'

2e-t) -- E ( - 1 ) " u=O

22ve_tv

"

(2k + 1)2~ = (2k + u + 1)... (2k + 2u) - (2k + u + 1)v. (2k + 1)~

So we get n-k

n-k

~=o

~=oA~ J e kt

v!(n- k - v)!

e

-vt

0

= - T -s by (2.17), which completes the proof.

D

Now, in the discussion preceding Theorem 2.9, (e) holds by definition, and (d) follows from Theorem 2.11 and a theorem of Askey and Gasper that depends on the hypergeometric function, that will not be proved here, but that asserts that m

> 0. v=0

As to (c), it is immediate from the definition of Tk(t) in (*). To prove (b), observe that p(2k,o)(-1) - (-1) ~ (see the explicit definition in (2.15)), so by Theorem 2.11 {9

n-k

O---~Tk(O)-- -k E p ( 2 k '~

n-k

-k E(-1) ~

v=O

_ {0 -k

v--0

ifn-kisodd if n - k is even.

7.2. Some Coefficient Theorems

291

But if (a) holds, then +

1 (0) k+l

)

1

_

in every case; so Tk (O) -- T k + l ( 0 ) - -

1

and so summing rk (0) - n + 1 - k. It remains to prove (a). T h e o r e m 2.12. With the definitions above -I -~ k+l

k+l

as stated in (a) in the discussion before Theorem 2.9. Proof. We need to show that 10

Tk(t) + -~-~Tk(t) -- "rk+, (t)

1

0

k + 1 -~rk+l (t)

and to do this, given the form of what we need to prove, it is useful to introduce exponential factors and to consider ,k(t)

a~(t)- ---~-e

k,

and

bk(t)-

,-k(t)

k e-

k,

Then from the definition of Tk(t), we have (for 1 _ k < n) n-k

ak(t) -- E (--1)" (2k + v + 1)~(2k + 2v + 2 ) n - k - , .=o

(k + . ) . ! ( ~ - k -

.)!

e

-pt

and n-k

bk(t) -- E (-1)v

(2k + u + 1).(2k + 2v + 2)n-k-~,

(k + v)v!(n - k - v)'

v---O

So

e

-vt-2kt

"

n-k

0 -~a~(t) - E(-1) ~=1

TM

(2k + u + 1),,(2k + 2u + 2)n-k-~, _,,, (k + u ) ( u - 1 ) ! ( n - k - u)! e

and

n-k (2k + u + 1)v(2k + 2v + 2)n-k-,, 0 bk(t) -- E ( 1)~+1 (2k + v)e -'t-2k+kt

ot

(k + ~)~'(n- k - ~)'

m

v=0

n-k

= E(-1) ~=0

"

~'+1

"

(2k + v)v+1 (2k + 2u + 2)n-k (k + ~ ) , ! ( n - k -

~)!

-~'

e

-vt-2kt

7. Bieberbach Conjecture

292 On using (2k + v)(2k + v + 1)~ - (2k 0 -~bk+l(t) :

n-k-1

E v'--O

+ v)v-t-1.

So we have

(-1)v+l (2k + 2 + v)v+l (2k + 2v + 4)n-k - 1-v e_Vt_2(k+l)t (k + 1 + v)v!(n - k - 1 - v)!

n-k =~(-1)~

(2kv + 1)~(2k + 2v + 2 ) n - k - v -vt-(2kT1)t (k+v)(v-1)!(n-k-~)i e

v--1

On replacing k by k + 1 and then v by v - 1, thus n-k

e_kt 0 -~ak(t) -- Z

v=l

(--

1)~+ ~ (2k + v + 1)~(2k + 2v + 2)n-k-~, -(k+~,)t (k + v ) ( v - 1 ) ! ( n - k - v)! e

: _e(k+l)t i) o bk+ i (t) . But ak(t) -- r~---~ekt and bk+l (t)

-

r kk+l +l(t)

e- (k+l)t so we have

1 0 rk(t) + -s -~r~(t) - rk+l (t)

1 0 k + 1 -~rk+l (t) Yl

which proves Theorem 2.11 and so (a).

This completes DeBranges proof of Milin's conjecture, except, of course, for the unproved statement (d) depending on Askey and Gasper's results, and, of course, for the motivation of DeBranges' choice of weight function 7k(t). It remains to see how Milin's conjecture implies Bieberbach's (especially since DeBranges methods cannot be used directly to prove the Bieberbach conjecture.) T h e o r e m 2.13. Let f E S. So f ( z ) = z + a2z 2 + a3z 3 + . . . conjecture, [an] <_ n is true.

then the Bieberbach

Pro@ Let g(z) be an odd function in S such that (g(z)) 2 = f ( z 2) on B(0, 1). So g ( z ) -- Z nt- C3 z3 "4-C5 z5 + . . .

Suppose h(z) = ~llog ~

- ~nC~=1%zn. If z E B(0, 1 ) - (-1, 0], then

g(vq)) But

g(v )

:

1 + c 3 z + c5 z2 -I- . . .

and so is analytic in B(0, 1) Thus h(z) 9 (with cl - 1) E n----i

z

89 log I~) is a branch of log g(J7) V~

C2n+I z n -- exp

~[n z n n--I

.

9

So

7.2. Some Coefficient Theorems

293

So, by Theorem 2.8,

IC2k+112< (n +

I) exp

k----0

klTk12

,I

n+l

-

1/}

m---1 k - 1

But, by Milin's conjecture proved in Theorems 2.9 and 2.10 and the discussion there, (see discussion following proof of Theorem 2.9), this means that ~k=olC2k+l 12 <_ n + 1, or replacing k by k - 1, and n by n - 1, ~ k = l IC2k-1[2 <-- n. This is, in fact, a conjecture made by M.S. Robertson in 1936, and it implies the truth of the Bieberbach conjecture. For consider f and g as above. Then f ( z 2) - z ~ + a2z 4 . . . .

(z + c3z 3 + . . . ) 2 .

Expanding and identifying coefficients, this gives for all n >__1 (with cl - 1)

an -- E C2r-lC2n-2r+l

9

r--1

So by Cauchy-Schwarz, n

r----1

by the above.

[-7

N o t e 2.8. There is a way to prove the inequality on Tk(t) independently of Jacobi polynomials. This is brief but still requires knowledge of generalized hypergeometric series. See the paper by Gasper [83]. It is, in fact, interesting that the critical fact in this proof comes down to an inequality on generalized hypergeometric series (and unfortunately omitted above). In the nearly seventy years between the posing of the Bieberbach conjecture and its final solution by DeBranges, much work went into various attempts to prove it, at least for some classes of univalent functions. Thus it was proved for typically real univalent functions and for univalent functions that mapped B(0, 1) onto a "star-shaped" region. These and similar results can be found in the first edition of this book. However, for some univalent functions, a sharper result than la,~l _< n has been known for a long time prior to DeBranges Theorem. We close this chapter with such a result. T h e o r e m 2.14. If f(z) - z + EnC~__2anz n E S, and the image T of B(O, 1) under f is convex, then lanl ~_ 1, and this result is sharp.

Proof. Let Tr be the image of B(0, r), 0 < r < 1, under f. Suppose f(zl), f(z2) e Tr, where zx,z2 E B(O, 1) and zl # z2, IZll _ ]z21. Then since T is convex, for all z E B(0, 1), and 0 < t < 1, F(z,t)-tf

( zl z

+ ( 1 - t)f(z) e T .

7. Bieberbach Conjecture

294

So writing f--1 for the inverse of f; f - l ( F ( z , t ) ) is analytic in B(0,1) and [f-X(F(z,t)l < 1, f-X(F(O,t)) - 0; and so by Schwarz's Lemma, lf-lF(z,t)! <_ Izl for all t E (0,1), and so in particular, taking z - z2,

If-l(t/(zx) + (1 - t)/(z2))! < Iz21 < r, for all t e (0,1). Hence the line joining zl and z2 is in T~; that is, Tr is convex, 0 < r < 1. So the curve { f ( z ) ' l z I - r} is convex for each r, 0 < r < 1. The tangent at the point re i~ makes the angle 7r/2 + 0 (measured counter-clockwise) with the real axis; hence the tangent to the point f(re i~ on the image curve makes the angle

co - 7r/2 + 0 + arg f'(rei~ . Now, if s denotes the arc length of the image curve, then, d8

>0.

dO

Also, ~d~ denotes curvature, and since T~ is convex (and 0 E T~), ~god is nonnegative. Hence

dw dO

dw ds >0 ds d O - "

--= That is

d ( T r / 2 + 0 + arg f'(rei~ dO

>0

1 + ff---~Im( Log f'(rei~

>_ O,

-

'

or

SO

l+Im

( irei~ f"(rei~ i~

- l + R e (Zf"(z))\f'(z)

_>0

with z - re i~ And this holds (by the maximum modulus principle for the real part) for all z E B(0, 1). It follows from this that

Re

z&(zy'(z))) > o zf'(z)

-

from which it easily follows that zf'(z) is univalent on B(0,1).

zf'(z)

- 0 and ~ ( z f ' ( z ) ) ~z=0

Since clearly

- 1, it follows that zf'(z) e S. But z=0 cx)

z +

n=2

and so lanl _< 1.

I-1

7.2. Some Coe~cient Theorems

295

N o t e 2.9. Before DeBranges' proof of the Bieberbach conjecture, this could be deduced from the fact that the image of B(0, 1) under zf'(z) was starshaped and R. Nevanlinna's proof in 1917 that for such functions la~l <__n.