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1, so that
=
< e,
and deduce
Ia 0. We have further
Theorem 3.2 Suppose that f(z) is mean p-valent in izi < 1 and set A = max(2, 22 ), when 2> 0. Then d
SAr, f) = r— lAr, f) pAM(r, f) ;- (0 < r < 1), dr
(3.9)
and f)
Mfro, f) À + pA
r
M(t, f) ldt
(0 2, we deduce, since W(R) > 0,
p(r, R)R 11-1 dR
M/1-2 W(M)
1-;-M
P M(r, f) A .
2
3.3 Estimates for the coefficients If 0
1) where A i (p,#) depends on p, )3 only.
We shall need the following preliminary result: Suppose that f(z) is mean p-valent in 1z1 < 1 and that r < 1, 0 < A. < 2. Then there exists p such that 2r — 1 < p < r
Lemma 3.1 1
and 1
2n
r2n
frl If(Pel° )1 2 1f(Pele ) 2-2 01 < 4PM(r .1(1 — r) '
(3.13)
We deduce from Theorems 3.1 and 3.2 that 27E f r pdp f It(pel° )1 2 1f (pe )1 2-2 de S(r) < M(r f)a - 2 A2 A Ljr 2 r — 1 0
Hence we can choose p so that 2r — 1 < p 1. Also since M(r, f) 15_ (1 — WI , En does not meet 1z1 < r if 2' > (1 — r) - 1, i.e. if
n> 1 --1- [
1 1 log ] = N(r), say. 2log 2 1—r
Hence
j 27 PaP of 1 +Ippeio)12 pm {1+
r
f
o
If'(pe`')12pdpc/0 N(r)
4
1
= mr[1+4N(r)],
and this yields (3.19). Suppose now first that fl = I-. We suppose that r > ri = 2r —1. Then using Theorem 3.2 and (3.19) we obtain
3 -4:'
and write
27-c
21ir fr: pdp fo It(pe` () )40} 2 {
--T.
On the other hand, we have for lz < 1, 00
11(z)-11
1
= E,
n=1
k=1
so that the series for f (z) converges uniformly and so f(z) is continuous in lz < 1. Also the area, with due count of multiplicity, of the image of lz < 1 by f(z) is
r 2ir rdr
Jo
co
co
If(re )1 dO = rt E ni a n I 2 '0 2
n=1
6'2 2 -2k = ne k= 1
2
3
(see §1.3). If 71W(R) denotes the amount of this area which lies over I w < R, then W(R) = 0 if R < 1— e W(R) < ire2 /3 otherwise. If we choose e < 1, it follows that ,
W (R) IT R2
7te 2 3702
4E2 3
(0 < R < co),
and we can make the right-hand side as small as we please by choosing 6' small enough. Thus f(z), which is continuous in f(z) < 1, can be made mean p-valent there with p as small as we please. Clearly, for any function f (z) bounded and mean p-valent in lz < 1, nlan 1 2 converges and so
Examples
77
(3.21) holds. Nevertheless (3.22) shows that nothing stronger than this need be true.
Examples
33
If f(
cc
an zn is mean p-valent in 1z1 < 1 and M(r, f) = o(1 —
, as r
1
where fl > 0, prove that 1,(r, f) = o(1 — r) 1-fi'l as r --+ 1, provided that f3> 1. Deduce that, if /3>
I i(r, f') = o(1 — r)-fi as n
co,
and
= o(n 3.4
1 ) as n --+ co.
(Use Theorem 3.2 and Lemma 3.1.) Suppose that g(R) is positive increasing for R > 0 and that
" RdR Jo g(R)
0, and g(0) = 2G"(0). If IG(r, f)
= 27cf0 27r ufrew )de,
78
Means and coefficients
deduce from Green's Theorem that if d SG(r, f) = G(r, f) then
sdr, f =
pr 2n Jo PdP 0f If f (Pei° )1 2
(Pe1° )140
Rg(R)p(r, R)dR. By writing G(R) = (e 2 ± R2 ,) 14 , and letting e tend to zero, deduce another proof of Theorem 3.1 (Flett [1954]).
3.5 Coefficients of general mean p-valent functions and 3.4 give immediately
Theorems 2.5, 3.3
an z n is mean p-valent in lz1 < 1.
Suppose that f(z) — Then we have for 1 < n < co
Theorem 3.5
< A(p),up n2P- ',
(3.23)
Suppose that f(z) c
r10(/1 ) = -- + A + (i — A + 4.12 )
S. Then if 0 < .3. < 1 and
we have
1,1(r, f') < 24(q, 2)(1 — r) -'
0 ._ r < 1.
(3.27)
— /1), we have
Hence if )6 > Po = {lo(A) + (1 —
I i (r,LI ) < AW)(1 — 0 -13 , f
< r < 1. 21 —
(3.28)
Choosing .1= .07 we obtain (3.28) with fib < fl < .4905525
.59944 of Theorem 3.6, corollary. The methods of Carleson and Jones are based on fractals and iterations and are unfortunately outside the scope of this book. Instead we shall in the next section develop a result of Baernstein [1986].
3.5.3 Baernstein's extension of Theorem 3.3
We proved in Section 3.5.1 < 1 and that (3.27) for the class S with constants ri, /1., where 0 < < P., yields (3.29), i.e. i(r,
— r) -1 +K
e7 (1 — r).
(3.45)
3.5 Coefficients of general mean p-valent functions
Lemma 3.5
L
87
With the above hypotheses we have 51 - 1C
Ifl(reie)Ide
._
A6 ( 1 - P
)
'
We suppose without loss of generality that 4) = 0, so that / is symmetrical about the real axis, since this may be achieved by a rotation. Thus z(/) = p. We now consider the bilinear map p 1 — pz Z -
C=
(3.46)
followed by the radial projection Z=R
C IF'
where R =
r—p 1— pr .
(3.47)
Writing r = 1 — E, p = 1 — ô, we have e < e-7 6 < -1- e-7 by (3.45). Thus R=
6—e 1— e 7 > > .998. (5 + E(1 — (5) 1 + e-7
Let El be the image of E in the ( plane and let e be the radial projection are related by (3.46). of E on 1z1 = R. We write f(z) = g((), where z, Then, since r> 1 by (3.45), we have
1
Inz)Ide -- 2 f Iff(z)IldzI = 2 E
f 1001141-
(3.48)
I
We note that for lz = r, we have z—p 1— pz
2
(1 — r 2 )(1 — p 2 ) 11 — pz1 2 =
Thus E l lies outside ICI = R. If
(1 — r2 )(1 — p 2 ) . (1 — pr) 2
(3.49)
= te' then, as z = rei° varies on /, we
have
4
1 dt P — + idip = dz { + t z —p 1 — pz J — 1 + p = izdO { z — p 1 — pz 1 1 1. = idO { P + z — p 1 — pz f
(3.50)
We take imaginary parts in (3.50) and write z = x+iy,1/1 = 2y. By (3.43), y < le-9 on /, and r>1— 4.y. Thus
x = rcost9 > (1 —4y)cosy > (1 —4y)(1 — ly)> 1— Sy > p
Means
88
and coefficients
since p =1— 2e8 y by (3.45). Thus — ' ox de. 1 1— PzI 2
dip = de I P(x — P) + 1— Px 1 > 1 1 lz — PI 2
1 1— PzI 2
We recall that p = 1— 6, r =1 —
E
so that y = e -8 6 by (3.44). Thus
11 —pz11—pd-plz-111—pd-(1—r)-1-101
ô1-+y < 1.016 (3.51)
and 1 — px _._ 1 — p = 6. Hence
dip
(5 1 dO > (1.012 > 1.036 .
(3.52)
On the other hand lz — PI
r—p=6—E> .996, 11 — pz I
1 — px 6.
Thus (3.50) yields 4
de
1 I 11 1+ 61
2.02 .9J
> 11 — pz1 2 — (1.01) 26 2 since r > .99, p> .63. Also (1 — r)(1 +p) 2(1 —r) 1— R = < and 1 1 — pr 1—p
R>
1— r 1—p •
(3.55)
Thus 3 (1 — t) > — (1 — R). 4
(3.56)
The function h(w) = g{Re iw + (1— R)w} is univalent in iwi < 1. Hence (1.3) in Theorem 1.3 yields for 114,1 < 1 1 + lwl lh / (0)1. Ik(w)1 ( 1 — Iw1) 3
3.5 Coefficients of general mean p-valent functions
89
We define w by
Re"" -1- (1 — R)w = = teiw, so that
t—R 1 1w1 = 1 — R < -4 by (3.56). Thus
5/4 ,
1001
Ig (Z)1 < 3 10Z)1.
On combining this with (3.48) and (3.54) we obtain
JE If' (Ode < A7 f (Re nP )Idtp, with A7 = 18. We apply Lemma 3.4 with g instead of f, R instead of r, g(0) = f(p) and M as before. Thus
If' (z)1(10 < A4A71f(P)1 27(2-)) A 1(2-2A)/(2-2 `) ( 1 — R)K— /. Using also (3.55) and K
0. We recall that /1, ri are constants such that (3.27) holds and K is defined by (3.36). Let k be the smallest nonnegative integer such that
2-k M < e8 (1 — r)-v.
(3.57)
We define
< 27r and 1f(re i° )1 2-k M}
Bk =
and, if k > 0 and 0 < j < k,
Bj = (010
2n, and 2-1-1 M
K —
If (rei6 )1d0 < A4(2- k m)(2-2i.)/(2-).) (1
1 by (3.37) with
Means
90
and coefficients
If 2-k M < 1, we deduce that
(re16 )40
A4(1 —
A4(1 —
If 2-k M > 1 we obtain
If(rele )Ide
< A4e8 (1 — r)-'"
A42 -k M(1 —
= A4e8 (1 — r)4 .
(3.58)
Thus (3.58) holds in all cases. If k = 0, Bk = [0, 2m] and (3.38) is proved. Thus we now suppose that k > 0, and consider Bi for 0 j < k. Hence 2-1 M > e8 (1 — r)'. We fix such a value of j and define
p = 4(1 — r)-'6 21 M -1 ,
(3.59)
Using also (3.37) with C = 1, we deduce that 4 X 2i < p < 4e-8 (1 — 0'13 = 4e-8 (1 — r) -".
(3.60)
Let 1 be the largest number of the form 2n /n, where n is a natural number, such that
1 (3.61) < —(4e-8 ) 1 /' < e 16 16 by (3.60) and since a < Thus (3.43) holds if I is an arc of length 1. By (3.61) we have n> 1. Thus 1 2m > — (1 — r)p1/2, (n — 1) 16 so that using (3.60) we have n — 1 27E (3.62) > 1 (1 r)p l /' > (1 — r). /= nn— 1 32 Thus (3.43) holds, when I/I = /. We divide the interval [0,2m] into n arcs I of length 1 = 2Tc/n and apply Lemma 3.5 with Ej = I n Bj. We recall that by (3.37), (3.44) and (3.62) 1
If (Pell
1 —(1 — r) 16
(1 — p )
( e / 8 ) fl < {32e 8 (1 — r) i r i/œ lfl •
Also by (3.61)
1—p 1— r
e8 1 1< 1— r
e8 1/, 16
3.5 Coefficients of general mean p-valent functions and
—
K
1, a
> .4,
and fl > a, we obtain 8)A
e 2 ( _i_i
21.3.2 > e 2 . =e
Now Theorem 2.4 yields no/log ( A°(1 P) ) < 1— r log
4 log(R2/Ri)'
2
(t)
—1
i.e. 4 log {
no < log (-/k) By (3.67) we have R2 10g (- ) > 2 + fi —œ logp > -1± logp,
a
while by (3.61) log p > a log
16e-8 (1 — 0) 16/ " > = a log 1—r 1—r
{log 1
P 6} .
3.5 Coefficients of general mean p-valent functions
93
On the other hand by (3.45) log
(1 1 Pr )
a, 1 -p t , > 7, so that log p> og 7
1-r
Thus V 1-p R2 v , log — > - log p > - log R1 a 7 1-r and
4 log = 11-re+4 log A0 < 28 ± 2 log Ao. V log /3-2log lk Ri
no
— b1 and so Ib i
4albo l, i.e. (1 —r2 )1f(ret ° )1
4alf(reie )1.
We deduce 0 4« — loglf (retc )1 (0 < r < 1), Or 1 — r2 and integrating with respect to r, we obtain
e
f(r ) < ( 1 + r f(0) 1 r —
f For generalizations to mean p-valent and other functions see Hayman [1952].
3.7 k-symmetric functions and Szegei's conjecture Since a
.491. Thus in (3.69) .245 cannot at any rate be replaced by .7> sin(.246n).
If fk (z) is regular in
3.7 k-symmetric functions and Szegii's conjecture IzI 1.
(3.71)
The result was obtained by Littlewood [1925] for k = 1, by Littlewood and Paley [1932] for k = 2, by V. I. Levin [1934] for k = 3 and by Baernstein [1986] for k = 4. Littlewood [1938] showed that (3.69) is false for large k and an example of Pommerenke [1967, and 1975, p. 133] proves that it is false for k > 12. The conjecture (3.71) is attributed by Levin [1934] to an oral communication of Gabor Szeg6. We note that since fk(z)= f(z k ) 1/k
we have M(r,f k ) = M(rk ,f ) 1 /1` < r/(1 — rk )2Ik
0< r .4 instead of fi > .491. This seems well beyond techniques available at present. If fk(z) is k-symmetric and p-valent a similar argument, based upon Theorem 3.3, shows that < A(p,k) ppn (2P 10-1
(3.72)
provided that 1 < k 4p, as is shown by a suitable form of the examples in §3.4. For this type of proof is essential that all the coefficients vanish, except those whose suffixes form an arithmetic progression of common difference
3.7 k-symmetric functions and Szeg6's conjecture
97
k. We proceed to prove (3.72) under somewhat weaker assumptions, 'basing ourselves on the fact that the functions fk(z) satisfy Ifk [re i(°+27" /k) ] = Ifk(re i8 )1 ( 0 < 3.7. 1
k —1 ).
We need the following preliminary result:
Theorem 3.10 Suppose that f(z) = and that there exist k points k > 2, such that (i)
E ooc an z" is mean p-valent in 1z1 < 1
on Izi = r, where 0 < r < 1 and
(1
i < j < k)
and
If(4)1
(ii)
R (1 < i
k).
Then we have R < A(p)0 2p(i/k—i) (1 _ r)-2p/k ,
where p is given by (2.8). We suppose that 6 > 4P+2(1 — r).
(3.73)
For if this is false, we have by Theorem 2.3
M(r, f) < A(p)p(1 — r) -2P 471_2
2p-2plk
A(p)p(1 — r)-2P/k (
6 ) < A(p)p(1 — r) — 2plk 6 2(p/k) -2p so that Theorem 3.10 holds. At least one of the annuli 1 — 4'6 < IzI Since M(r, f) increases with r, the result for 0 < r < -I also follows, and (3.78) follows from (3.77) and Theorem 3.3. Thus Theorem 3.12 is proved. It is worth noting that the above argument does not require the full strength of our hypotheses. It would be sufficient to assume that on
1z1 = r cc
= E abm+,,L,bm+v = 0(1 — r)-P
as r -- 1,
m=0
in order to obtain M(r, f) = 0(1 — r) -P. A similar remark applies to Theorem 3.11. The conclusion of Theorem 3.11 continues to hold if an = 0, except for a sequence nv such that n +1 — n, > k finally (instead of n„ +1 — n,, = k finally) [Hayman 1967]. Similarly that of Theorem 3.12 holds if an = 0 for a sequence n = nv , such that n„ +1 — n, < b [Hayman 1969]. These results involve some Fourier series arguments to show that the hypotheses of Theorem 3.10 are satisfied and are beyond the scope of this book.
Examples
3.6
Show that if f(z) is univalent and satisfies the hypotheses of Theorem 3.11 for k = 4, then Ia n ' < A(N)tn. (Use Theorem 3.7.) Note that this conclusion extends Szegb's conjecture for
k = 4.
102 3.7 3.8
Means and coefficients
If f(z) is univalent as well as mean p-valent in Theorem 3.12, show that (3.78) continues to hold for p > .491. If k > 4p in Theorem 3.11, or p < -1 in Theorem 3.12, show that i a, = o(n) and that this conclusion is sharp. (Use Theorem 3.4 and the example of Section 3.4.)
4 Symmetrization
4.0 Introduction
In this chapter we develop the theory of symmetrization in the form due to 1361ya and Szegei [1951] as far as it is necessary for our function-theoretic applications. Given a domain D, we can, by certain types of lateral displacement called symmetrization, transform D into a new domain D* having some aspects of symmetry. The precise definition will be given in §4.5. 1361ya and Szeg6 showed that while area for instance remains invariant under symmetrization, various domain constants such as capacity, inner radius, principal frequency, torsional rigidity, etc., behave in a monotonic manner. We shall here prove this result for the first two of these concepts in order to deduce Theorem 4.9, the principle of symmetrization. If f(z)= ao + aiz +... is regular in 1z1 < 1, and something is known about the domain Di- of values assumed by f(z), this principle allows us to assert that in certain circumstances lad will be maximal when f(z) is univalent and Di- symmetrical. Applications of this result will be given in Sections 4.10-4.12. Some of these will in turn form the basis of further studies of p-valent functions in Chapter 5. Some of these results can also be proved in another manner by a consideration of the transfinite diameter (Hayman [1951]). The chapter ends with a recent proof of Bloch's Theorem by Bonk [1990]. Since the early part of the chapter is used only to prove Theorem 4.9, the reader who is prepared to take this result for granted on a first approach may start with §4.5 and then go straight to §4.9. We shall need to refer to Ahlfors [1979] (which we continue to denote by C. A.) for a number of results which space does not permit us to consider in more detail here. A good set-theoretic background is provided by C. A., Chapter 3, §1, and we use generally the notation there given.
103
104
Symmetrization
We shall, however, in accordance with common English usage, call an open connected set a domain and not a region. The closure of a domain D will be denoted by b. A domain D has connectivity n if its complement in the extended plane has exactly n components; if n = 1, D is simply connected, if n = 2 doubly connected, etc. (C. A. pp. 139 and 146). If the boundary of D consists of a finite number n of analytic simple closed curves (C. A. pp. 68 and 234), no two of which have common points, we shall call D an analytic domain. We shall assume a right-handed system OX,OY of rectangular Cartesian axes in the plane. Points in the plane will be denoted in terms of their coordinates x, y either by (x, y) or by z = x iy, whichever is more convenient. Accordingly, functions u will be written as either u(z) or u(x, y).
4.1 Lipschitzian functions Let E be a plane set and let P(z) be a function defined on E. We shall say that P(z) is Lipschitzian or Lip on E if there is a constant C such that
P(z2)
— z21
(4.1)
whenever z2 lie in E. It is clear that a Lip function is continuous, and further that if P,Q are Lip and bounded on E, then PQ is Lip on E. Suppose that E is compact and that P(z) is Lip in some neighbourhood of every point zo of E. Then P is Lip on E. For if not, we could find sequences of distinct pairs of points z n , z n' (n > 1) on E such that
IP(zn )— P(4)1 1zn — 41
00
(n —> co).
(4.2)
zo , By taking subsequences if necessary we may assume that z n z z'o , where zo, 4 lie in E. If zo, 4 are distinct, we at once obtain a contradiction from our local hypothesis and (4.2), since P is bounded near any point of E. If zo = 4, then zn, z'n finally lie in that neighbourhood of zo where P is Lip and this again contradicts (4.2). If P(z) = P(x, y) is defined in a disc y (C. A. p. 52) and has bounded partial derivatives there, then P(x, y) is Lip in y. Suppose first that P(x,y) is real. If (xo,yo), (xo +h,yo +k) both lie in y, then either (xo, yo + k ) or (xo + h, yo) also lies in y. Suppose, for example, the former. Then if M is a bound for the absolute values of the partial derivatives in y we have from the mean-value theorem
4.2 The formulae of Gauss and Green
105
IP(x0 + h, yo + k)
P(xo, Yo)I IP(xo + h, yo + k) — P(xo, yo + k)I + IP(xo, yo + k) — P(xo, LP)
h
(
Ox (x o +Oh, yo+k)
M(1111+1k1)
k " \ °Y
(x 0 ,y0+6q)
2M../(h2 + k2 ).
Thus P is Lip in 7. For complex P we can prove the corresponding result by considering real and imaginary parts. If follows that if P has continuous partial derivatives in a domain containing a compact set E, then P is Lip on E. For in this case P has continuous partial derivatives in some neighbourhood and so bounded partial derivatives in some smaller neighbourhood of every point of E. Conversely, we note that if P(x, y) is Lip on a segment a < y < b of the line x = constant, then P is an absolutely continuous function of y on this segment and so OP/Oy exists almost everywhere on the segment and is uniformly bounded. t Thus
P(x, b)— P(x, a) =
OP(x, y) dy. a OY
(4.3)
4.2 The formulae of Gauss and Green We proceed to prove these formulae in the form in which we shall require them in the sequel.
Lemma 4.1 (Gauss formula) Suppose that D is a bounded analytic domain in the plane and that its boundary y is described so as to leave D on the left. Then if lc, y), Q(x, y) are Lip in b, we have
OQ OP (Pdx + Qdy) = f f (— — —) dxdy. D °X OY Let z = a(t) (u < t < b) give an arc of y. Then a(t) is a regular function of t and i(t) 0. The tangent is parallel to OY at those points where ce(t) is pure imaginary, and this can be true only at a finite number of points, since otherwise a'(t) would be identically pure imaginary and so this arc of y would reduce to a straight line. t This is, however, impossible, since y consists of a finite number of analytic closed curves. Thus there are only a finite number of tangents to y which are parallel to 0 Y, and Burkill [1951], Chapter IV. The real part of a' (t) is a regular function of t for a < t < b and so has only isolated zeros or vanishes identically.
106
Symmetrization
we assume that these are
x = xm (1 m < M), where x1 <X2 < < xm . For xm_i < < xm, the line x = ç meets y in 2n points Y = Yi(),
-•
Y = Y2n(),
where n depends only on m and the y v ( ) (1 < y < 2n) are differentiable functions of ç for xm_i < < xm . Also the part Dm of D lying in xm_ i <X < xm consists of n domains Dm,v
< x < xm (1 < V < n),
y2 v- 1 (x ) < Y < Y2v(x),
since at the intersections of x = and leaves D. Consider now
with y, the line x =
alternately enters
I = f P(x, y)dx taken along y so as to keep D on the left. Then dx > 0 on the curves y = y2v-i(x) and dx < 0 on the curves y = Y2v(x). Thus if /m is the integral taken over those points of y which lie in xm_ i < X < xm, then xm n
Im =
El/3[x, y 2v_,(x)] _ P[x, y2 v (x)]}dx
f
Xm-1 v =
1
Xm
=
—f
dx
E= 1 v
ry2v
OP(x,y)dy ay
_OP
L-1 v=1
.1
dxdy
CY
by (4.3), since P(x, y) is Lip. Adding over the separate ranges xm_i <x < xm and all the domains Dmo, we obtain OP
Pdx = f f — dxdy. D
CY
Similarly we prove
aQ 1Qdy = f f dxdy, D OX
and Lemma 4.1 follows. We deduce Lemma 4.2 (Green's formula) Suppose that D is a bounded analytic domain with boundary y, that u is Lip in b, and that y possesses continuous
4.3 Harmonic functions and the problem of Dirichlet
107
second partial derivatives near every point of b. Then
,,-))1
a2v ) 4_ (P_IL1 -!-) 4- 'ill- -°2 u °v ds = — f f [--u °21j ( ax24_, oy 2 ' o x Ox Oy cy j j dxdy, D fy where a/an denotes differentiation along the normal into D, and ds denotes arc length along y. In fact u, avgx, av/ay are Lip in b in this case and hence so are u(av lax), u(av/ay). Let (xo, Yo) be a point of y and let 0 be the angle the tangent to y makes with OX, where y is described so as to keep D on the left. Then the normal into D makes an angle 0 + -In with OX and
Ov -an
lim h-+0
v [xo + h cos(6 + i'r), yo + h sin(0 + -1 Tc)] — v [xo, Yoi h
av . av = cos 0- — sin 0—. 0y Ox If ds is an element of length of y at (xo, RI), its projections on OX, 0 Y are dx = ds cos 0 and dy = ds sin O. Thus
0v
av
av
— ds = —dx — — dy. an ay ax We now apply Lemma 4.1 with P = u(av/ay), Q = —u(av/ax) and Lemma 4.2 follows.
4.3 Harmonic functions and the problem of Dirichlet A function u(x, y) is harmonic in a domain D if u has continuous second partial derivatives in D which satisfy Laplace's equation 02 u
OX2
02u 2
ay2
It follows from this that (1)(z) = Ou/ax — i(au/ay) has continuous partial derivatives which satisfy the Cauchy-Riemann equations, and so 4)(z) is regular in D. Also in any disc with centre zo that lies in D, u is the real part of the regular function
z f (z) =
f70
(/)(C)dC + u(z0).
Thus u possesses continuous partial derivatives of all orders in D. Also, by the maximum modulus theorem applied to exp{+f (z)}, u can have no local maximum or minimum in D unless u is constant in D. Suppose now that D is a domain in the open plane and that u(C) is
108
Symmetrization
a continuous function of , defined on the boundary y of D considered in the extended plane. Thus, if D is unbounded, y includes the point at infinity. The problem of Dirichlet consists in finding a function u(z) harmonic in D, continuous in b and coinciding with u(c;) on y. If ul, u2 are two functions satisfying these conditions, then u 1 — u2 is harmonic in D, continuous in b and zero on y, and so by the maximum principle /4 1 — /42 vanishes identically in D. Thus a solution to the problem of Dirichlet is unique, if it exists. A solution does not always exist. If D consists of the annulus 0 < 1z1 < 1, then any functions bounded and harmonic in D can be extended to be harmonic also at z = 0 and so in 1z1 < 1. Thus boundary values assigned on 1z1 = 1 determine u(0). The question of existence can often be settled by means of the following result, for whose proof we must refer the reader to C. A. p. 250. Theorem 4.1 Suppose that for every boundary point z o of D there exists w(z), harmonic in D, continuous in b, and positive in I), except at z o , where co(z o ) = 0. Then the problem of Dirichlet possesses a solution for any assigned continuous boundary values on the boundary of D. From this we deduce the following criterion:
Theorem 4.2 The problem of Dirichlet always possesses a solution for the domain D if, given any boundary point of b, there exists an arc of 0 straight line or circle containing zo and lying outside D. We shall deduce Theorem 4.2 from Theorem 4.1. Suppose first that zo is finite. Let the arc c have end-points z o and z i . The function = ei
a Z — Zo z1 — z
maps the exterior of this arc c onto the plane cut along a ray through the origin, so that z = zo , z 1 correspond to = 0, co. By suitably choosing oc, we may arrange that the ray is the negative real axis. Then
0 —1
w= 1 + i
V+1
maps this cut plane onto the circle lw — 11 < 1. Also z = zo corresponds to w = 0, and D maps onto a domain whose closure lies in lw — 11 ._-. 1, and such that only z = zo correspond to w = 0. Thus co = 9iw is the
4.4 The Dirichlet integral and capacity
109
harmonic function whose existence is required in Theorem 4.1.t If zo is infinite and some ray from a finite point to infinity lies outside D, the argument is similar. This completes the deduction. A plane domain satisfying the criterion of Theorem 4.2 will be called admissible (for the problem of Dirichlet).
4.4 The Dirichlet integral and capacity
Historically Dirichlet attempted to base a proof of the existence of a solution to his problem on the problem of finding the minimum for the Dirichlet integral
= f fp [(4 1x4 ) 2 + (4) 1 2
ID(U)
dxdy,
for all functions having the assigned boundary values and satisfying certain smoothness conditions. In favourable circumstances this minimum is attained by the required harmonic function. However, even for continuous boundary values on the unit circle, the harmonic function inside the circle with these boundary values may have an infinite Dirichlet integral, so that the minimum problem has no solution. We shall need the Dirichlet minimum principle only in a special case, where its validity can be proved without difficulty.
Theorem 4.3 Let D be an admissible domain in the open plane whose complement consists of a compact set E l and a closed unbounded set E0, not meeting E l . Let v(z), co(z) both be continuous in the extended plane, 0,1 on E0, E i respectively and Lip on every compact subset of D. Suppose, further, that co(z) is harmonic in D. Then D [V(Z)1
D [W(Z)] Ya
On
ds,
where ya is the set {z : co(z) = a}, and a is any number such that 0 < a < 1 and Oco/ Ox—i(Oco/ y) # 0 on ya . In this case ya consists of a finite number of analytic Jordan curves. The system consisting of the domain D and the sets E0, E 1 will be called a condenser, and 'D [W(Z)] will be called the capacity of the condenser for evident physical reasons. In the special case when E0, E1 are continua, so that D is doubly connected, a simple function-theoretic interpretation t A point of b on the arc through ZØ,
zi corresponds to a pair of complex conjugate values of w. Thus co is uniquely defined and continuous even on this arc.
110
Symmetrization
is possible. We may then map D (1,1) conformally onto an annulus Atz : 1 < 1z1 < RI. (For a construction of the mapping function in terms of w(z) see C. A. p. 255.) The Dirichlet integral is clearly invariant under this transformation so that A has the same capacity as D. The harmonic function satisfying our boundary-value problem in A is
fl(z) =
log1R / z1 in A, log R
and hence the capacity of A is 2ir 1 pd0 1, i=i, log R p = log R . The quantity log R is frequently called the modulus of the doubly connected domain D. It is thus a multiple of the reciprocal of the capacity of D considered as a condenser. For our purposes it is, however, essential that E0, E1 need not be connected.
4.4.1 Proof of Theorem 4.3 Let w(z) be the function of Theorem 4.3. Then (/)(z) = aw/Ox — i(Oco I ay) is regular and not identically zero in D, and so (/)(z) vanishes at most on a countable set of points in D, which we shall call the branch points of co(z). Suppose that 0 < a < 1 and that the set Va, where co(z) = a, does not pass through any branch point of co(z). Clearly Va is compact and lies in D by our hypothesis. Also, near any point z0 of y a, CO is the real part of a regular function
w = f (z) = co + iwi = a + ib + ai(z — zo) + . . . . Here a l # 0, since 4)(z) * 0 on Va. Thus the inverse function z = f 1 (w) is also regular and the part of Va near zo is the image by a regular function of a straight line segment, i.e. an analytic Jordan arc. If we continue along this arc we must finally return to our starting point, since otherwise Va would posseses a limit point not of the kind described above. Thus Va consists of analytic Jordan curves, no two of which have common points. There can only be a finite number of these curves, since otherwise there would be a point of Va any neighbourhood of which meets an infinite number of curves. This again contradicts the local behaviour of Va established above. Suppose again that Va, Yb contain no branch points, that 0 < a < b < 1, and that Do = {Z : a < w(z) < b } . Then Do consists of a finite number of bounded analytic domains, of which Va, Yb form the complete boundary. Also if 010n denotes differentiation along the normal into D a, b
4.4 The Dirichlet integral and capacity
111
then Oco/On> 0 on ya and Ow/en < O on yb. Now, on applying Green's formula to the regions of Dad, separately and adding, we obtain aw n ds =0, f T aw n ds+ f T
I
a
ib
and since Ow/en has constant sign on Ow iYa
y a ,yb,
Ow
ds =
an
an
Yb
we deduce ds = I,
say. Also a second use of Green's formula gives aw ds— f w— aw ds = IDb [co(z)]. (b — a)I = — f w—
an
Ya
an
Yb
Since every point of D belongs to some Do, we deduce, on making a tend to 0 and b tend to 1 that
lim
D[W(Z)1 =
D a,b [W(Z)1
a—).0,b—■ 1
=1.
Suppose now that v(z) is the function of Theorem 4.3 and set h = v—w. Then h(z) is continuous in the plane and vanishes on E0, E 1 and so at infinity. Hence, given e > 0, the set
el
E = {z :1h(z)1
is a compact subset of D. Let ao, bo be the lower and upper bounds of w(z) on E, so that 0< ao 0, b —> 1).
Also 1Da, b ( V
)=
b
(
a)
)
b
(h)
21 Da b ( h,
Making a tend to 0, b tend to 1 we deduce /D(v) = /D(w) + /D (h). This proves the inequality of Theorem 4.3 and completes the proof of that
112
Symmetrization
theorem. We note also that equality is possible only if /D (h) = O. In this case it is not difficult to see that h(z) vanishes identically so that v(z) must coincide with w(z). We shall not, however, make any use of this.
4.4.2 Capacity decreases with expanding domain
We make an immediate deduction from Theorem 4.3, which has independent interest:
4,
E be pairs of closed sets characterizing Theorem 4.4 Let E0, El; two condensers as in Theorem 4.3. Let I, I' be their capacities and suppose that Eo c E0, Ei c El. Then I' __ I. Let w(z) be the potential function corresponding to the first condenser as in Theorem 4.3. If 0 < a < b < 1, we define a function v(z) as follows: v(z) = 0 where co(z) ._ a, v(z) = 1 where co(z) > b, and v(z) =
co(z) — a . , in Va,bb— a
4,
E respectively. Thus if Then v(z) is Lip in the plane and y = 0, 1 on D' is the complement of Ec,' and 4 Theorem 4.3 gives I I' < 'DO) --= — f - an s = 1 I D[w(z)] = d b — a. b— a ya, Yb
Making a tend to 0, b tend to 1, we obtain I' < I as required.
4.5 Symmetrization We shall now consider the process of symmetrization introduced in the middle of last century by Steiner and developed by 13■51ya and Szegii [1951]. Let 0 be an arbitrary open set in the open plane. We shall define a symmetrized set 0* in two ways as follows.
4.5.1 Steiner symmetrization
In this case we symmetrize with respect to a straight line, which we can take to be the x axis of Cartesian coordinates. For any real ç, the line x = ç meets 0 in a set of Mutually disjoint open intervals of total length l( ) say, where 0 _-_ l( ) __ co. The symmetrized set is to be the set
o. = {(x, y)
:
II < -/(x)}.
4.5 Symmetrization
113
Fig. 3. Steiner symmetrization In this case we symmetrize with respect to a half-line or ray, which we take to be the line 0 = 0 of polar coordinates r, O. We define the symmetrized set 0* as follows. Consider the intersection of 0 with the circle r = p, 0 < p < co. If this intersection includes the whole circle or is null, then the intersection of 0* with r = p is also to include the whole circle, or to be null respectively. Otherwise let 0 meet the circle r = p in a set of open arcs of total length pl(p), where 0 < l(p) < 2n. Then 0* is to meet the circle r = p in the single arc 4.5.2 Circular (Nlya) symmetrization t
1 0 1 < ]1 (P). We proceed to discuss some properties of symmetrization. Unless the contrary is stated results are true for either kind of symmetrization.
4.5.3 The symmetrized set 0* is open
We prove this only for circular symmetrization. The proof for Steiner symmetrization is similar. Suppose that (po, it) lies in 0*. Then 0 contains the whole circle r = po . Since 0 is open, 0 therefore contains some annulus po — 6 < r < Po + 6 and this annulus also lies in 0*. Thus (po, 7r) is an interior point of 0*. A slightly modified argument applies if the origin lies in 0*. ,
t
Nlya [1950]
114
Symmetrization
d
Fig.
4.
Circular symmetrization
Suppose next that (po, 00) lies in 0*, with 0 < po 21001. Choose x so that 21001 <x < l(p0 ). Then we can find a finite number of open arcs of total length greater than pox on the circle r = po and in 0. By diminishing these arcs slightly, we may take them to be closed and still lying in 0 and of total length greater than pox. If av < 0 < fi, is such an arc and 6, is its distance from the complement of 0, let 6 be the smallest of the 6,. Then the arc cx, < 0 < fl, of every circle r = p for Po — (5 < P < Po ± 6
lies in 0. Thus
l(p) _- x > 2100 1 for po — 6 < p < po+ 6, and so (r, 0) lies in 0* for
Po
—
6 < r < po d- 6 and 01< -1- x,
4.5 Symmetrization
115
where Ix > 100 1. Thus (p0,00 ) is an interior point of 0* and so 0* is open.
4.5.4 The Steiner symmetrized
set D* of a domain D is either the whole plane, or a simply connected admissible domain Suppose that the lines .x = Ç1, 2, where i < 2, meet D* and so D. Then x = meets D for i < < 2, since otherwise D could be expressed as the union of the disjoint non-null open subsets lying in the half-planes x < and x> respectively. Thus the segment < x < 2 of the real axis lies in D* in this case. Two points ( 1 , n i ), (2, 172) in D* can be joined in D* along the polygonal arc (1, qi), (2, 0 ), (2, 0), (2, 112). Thus D* is a domain. Suppose next that D* is not the whole plane, so that V) < +co for at least one . Then the complement of D* meets x = ç in the pair of rays ly1 > 1-1( ) . Thus the condition for admissibility in Theorem 4.2 is satisfied for every point (, n) in the complement of D* and also at co. Further, every point in the complement of D* can be joined to infinity by a ray, and so this complement is connected in the extended plane. Thus D* is simply connected and is admissible. The results for circular symmetrization are not so simple. If, for instance, D consists of an annulus pi po, 0 = it. Any boundary point of D* on this ray and the point at infinity clearly satisfy the criterion of admissibility. If (p,00) is any other point in the complement of D*, then p > po, 1 0 01 < it, and the arc IN! < 0 _-_ it of
116
Symmetrization
the circle r = p lies in the complement of D* and joins (p, 00) to the ray 0 = 7r, r _-_ po in this complement. Thus the complement of D* is connected and D* is simply connected. The criterion of admissibility is satisfied in all cases for every boundary point of D* not on the line 0 = 7r. The above argument shows that if D* is simply connected the criterion is also satisfied for boundary points (p, 7r). It remains to show that if D is an admissible domain, so is D*. By the above argument it suffices to establish the criterion of Theorem 4.2 for the boundary points (p0, 7r) of D*, and for co. Since D is admissible the complement of D contains a ray, which must meet every sufficiently large circle r = p. Thus (p, 7r) lies in the complement of D* for large p, and so this complement contains a ray r > p, 0 = 7r. Thus our criterion is satisfied at cc. Suppose next that (po, 7r) is a boundary point of D*. If an arc a < 0 < )6 of r = po lies in the complement of D, then an arc of r = po bisected by (po, 7r) lies in the complement of D* and the criterion is satisfied. If not, let (po, 0) be a frontier point of D. Since D is admissible there is an arc of a straight line or circle containing (po, 0) but lying outside D and so by our hypothesis not on r = po. This arc must meet r = p for all p either in some interval po — 6 ._ p _._ po or Po •-- P po +6. Thus the interval [Po — (5, Po] or [Po, Po + 6] of 0 = it lies in the complement of D* and contains (po, it), and so D* is admissible.
Let u(z) be a function, real, continuous and bounded in the plane. We symmetrize u to obtain a new function u*(z) by simultaneously symmetrizing all the sets 4.6 Symmetrization of functions
Da = { z : u(z) > a}(—cc < a < +co).
These sets are open since u is continuous. More precisely, let Da* be the symmetrized set of Da with respect to some straight line or ray. For any point z in the plane we define u*(z) as the least upper bound of all a for which z lies in D. In practice we shall be concerned only with the case where u(z) is nonnegative and vanishes continuously at infinity. In this case the sets Da are bounded for a > 0 and their closures Da are compact. Now a function u(z) continuous on a compact set E is uniformly continuous on E (C. A. p. 65). In other words if f2(6) is the upper bound of lu(z i ) — u(z2)1 for z 1 , z2 on E and 1z1 — z21 < (5, then Q( (5 ) is finite and Q( (5 ) —> 0 as (5 --- O. The quantity Q( (5 ) is called the modulus of continuity of u(z). Clearly u(z) is Lip on E if and only if i/((5) < C(5 for some positive C and 0 < (5 t. Suppose that
118
Symmetrization
b — fl(6) and jx2 — xi i _._ 6. Then
a _-_ t2 < ti —S1(6)
1(x2, t2) .1(xi, 0+106 2 —
(X2 - X1) 21
Let F1 be the set of all y such that u(xi, Y) > ti and let F2 be the set of y for which u(x2, Y) > t2. Then Y2 E F2 if yi E F1 and (x2
—
xi)2 + (V2
For otherwise we could find, on the line joining (x i , Yi) and (X2, Y2), two points at distance at most 6 from each other at which the continuous function u takes the values ti, t2 respectively, and this would contradict the definition of Q(6). Taking y2 = Yi we deduce at once that F2 contains F1. Further, let y' be the upper bound of F1 . Then u(x i , y') = ti . Hence it follows that u(x2, y) > t2 if (Y
—
02 + (x2
—
xi)2
and in particular y E F2 if y' __ y __ y' + V[6 2 — (x2 — xi) 21.
Similarly, if y" is the lower bound of F1 , y E F2 if y” — .06 2 - (X2 - X1) 2 1 --- y
y".
Thus F2 contains the whole of F1 together with two intervals, each of length V[62—(x2—x1)2], which do not belong to Fi. Since /(xi, ti), i(x2, 1.2) are the lengths of F1 , F2 respectively, the lemma follows. Suppose now that Theorem 4.5 is false. Then we can find (X19 Y1), (.X2, y2 ) on E*, such that for some positive (5 (x2
—
x1)2 + (y2 - Y1) 2 < 62,
and
a
u* (x2, y2) < u * (xi, Yi) — n(6) _ O.
Setting z = F(w), we see that z decreases from -?/-3- to 0, as w increases from 0 to 1, while w=
1 — z .\/3 1 — z/ \/3'
G(w)
1 — z \/3 (1 — z/ \/3) 3.
This yields (4.16) when z > 0. If z = teiŒ, we apply the conclusion to f (z el') instead of f (z). This completes the proof of (4.16).
4.13 Bloch's Theorem
139
We deduce that, with the hypotheses of Lemma 4.4, f(z) is univalent 1/0. For if zi, z2 are two distinct points in 1z1 1/0, write in 1z1 = z1 + pel'. The segment
0< t 0 in T. Thus
Z2 (z2)
-
f (zi)} =
fz i
f'(z)dz
f t(Z1+ tei')dt > 0.
=
(4.19)
Thus f(z) maps 1z1 < 1/V3 (1,1) conformally onto a domain D. The boundary OD of D is the image by f(z) of Izi = 1/V3. w = f ((1/ V3)el is a point of this image, (4.16) and (4.19) yield
1w1
w = f
.1/3
f(te `Œ)dt >
f \/3
(1 — tV3)dt
0 (1 -
0
t/.0)3
If
\/3 4
Thus D contains the disc lwl < and Lemma 4.4 is proved and so is Theorem 4.18 if f(z) E 00Suppose next that f(z) is not in 00. Then there exists zo, such that
(1-141 2 )1.04)i > 1 somewhere in 1z1 < 1. We choose p such that 0 < p < 1, and
(1 — 1zo1 2 )Plf(Pzo)1 > 1. This is possible by continuity. We now define tp(z) = f(pz) and
1 . (4.20)
p(z)= ( 1 — 1z1 2 )1V(z)1= ( 1 — 1z1 2 )P 1.r(Pz)1, 1z1
Then it(z) = 0 for 1z1 = 1, since p < 1, and so p(z) attains its maximum value po in 1z1_-_ 1 at z i where Izil < 1. Also po p(zo) > 1. We suppose that arg ip/(z i ) = a and consider
+
0(z) =
(-±
z
1)} •
We note that (1)(z) is regular in 1z1< 1 and 4)(0) = 0,
0'(0) =
e-12 ( 1 — 1z11 2 )17P'(zi) PO
1.
140
Symmetrization
Also we have for 1z1 < 1 (
1
-
1z 1 2 )10z)1 =
(1— 1z1 2 ) PO
+ z ) 1 — iz 1 1 2
1
Y)
2 =
1
1 + f1Z
ip
1
11 + f IZ1 2
-
+ Z1Z
(1+
,
( Z1 +
1 + fiZ
\/3/4. This completes the proof of Theorem 4.18.
4.14 Some other results We close the chapter by mentioning some further developments. Baernstein [1975] has proved that, with the hypotheses of Theorem 4.9, we have further I Ar, f)
I ( r, 0),
0
r < 1,
0 < 2. < oo
(4.21)
where
1/). 1 f 2ir , if I;.(r, f) = { 7, 0 If(rel 0 )r} and 1(r, f) = M(r, f). If f(z) E S he proved that I(r, f) -_ 1,1 (r, k), where k(z) is the Koebe function (1.1). Next Weitsman [1986] proved that Theorem 4.9 extends to arbitrary domains D, whose complement in the extended plane contains at least 3 points. If D* = Do and D* is multiply connected, 4)(z) maps 1z1 < 1 onto the universal covering surface over Do . It is not known whether the analogue of (4.21) holds in this more general case. The case when the complement of D* consists of the points (0, —1, cc) is of particular interest, since it leads to sharp forms of the theorems of Landau [1904] and Schottky [1904]. This special case had previously been dealt with by Hempel [1979] and Lai [1979]. For proofs of the above results the interested reader is referred to the original papers. A connected account is also given in Hayman [1989, Chapter 9].
141
Examples
Examples If 4.2
(1+W) 2 1-1,1, )
( 1 — r) 2 1 + r)
K 7 1+Z 2 1— z ) '
show that the C plane cut along the negative real axis corresponds (1,1) conformally to lz I < 1, and to lwl < 1 cut from —1 to —r along the real axis. Hence complete the proof of Theorem 4.14, first when M = 1, and then generally by considering f(z)/M, RIM, 1 instead of f(z), R, M respectively. The following examples are generalisations of Koebe's Theorem 1.2.
Examples
4.3
4.4
We denote by Wi the class of functions meromorphic in 1z1 < 1 and having s an expression (4.14) near z = O. Show that if f G 9J/ and a, b are positive numbers such that 1/a — 1 /b > 4, then D1 contains a circle 114,1 = r, such that a < r < b, (Hayman [1951]). Show by an example that this result fails whenever 1/a — 1/b = 4. (Assume the result is false and apply Theorem 4.9.) If w = f(z) E 9j1 and 0 < r < 1, show that D1 meets lwl = r in a set of linear measure at least 2rŒ, where r = cos (a/2), and for each r give an example for which this lower bound is attained. (Show that the map
z=
4.5
11— weli — (1 — w e —i')1 (e--11 — w)1 — (eil — w)
maps the complement in the closed w plane of the arc w = e 0 , a < 101 7E, onto 1z1 < 1.) If W = f(z) G Z where S is the class of Chapter 1, prove that, for 1 < r < 1, D = Df meets 1w1 = r in a set of linear measure at least 2rŒ where r = cos4(a/4). (If the conclusion is false, the complement of D* contains the ray —oo, —r of the real axis as well as the arc a < i arg z1 < it of lw 1 = r. To find the extremal map, set z Z =4k (1 +Z)2 ' (1 + z) 2 where k=
cos 3 (1 + cos i)2'
142
4.6
Symmetrization and W = rw,w , = f(z), is the map of Example 4.4. (Netanyahu [1969] has shown that a function of this type yields the extremal value 1a2 1 d = 2/3 where d is the radius of the largest disc with centre 0 contained in the image of lz1 < 1). If w = f(z) E Ti and f(z) is regular in lz1 < 1, show that for 0 < r < 1, Di meets 1w1 = r in a set of linear measure at least 2ra, where a is the (unique) solution, in 0 < a < 7r, of
r=
sin2 -1 2 log (sec -'21 ) '
This example needs the full strength of Weitsman's Theorem. The symmetrized domain D* lies in the domain Do, whose complement in the closed plane consists of the arc w = re'9 , a < WI < it and the point at oc. Thus, by Weitsman's Theorem the extremal map is w = ip(z) = r A i z+..., where tp(z) maps lz1 < 1 onto the universal covering surface over this doubly connected domain D o . In the map w = f(z) of Example 4.4, 1z1 < 1, punctured at z = c, corresponds to Do (1,1) conformally, and z = c = cos(a/2) is mapped onto w = cc. We now set
Z =
1 log z i + K c—z , K = log — . where z 1 = c log z i — K ' 1— cz
Then 1Z1 < 1 is mapped onto the universal covering surface of the unit disc 1z1 < 1 punctured at z = c. Thus the combined map w = ip(Z) = f(z(Z)) gives the required extremal map with tp(0) = 0,
Ilp'(0)1
r 2 log 1/c . 2r log 1/c = c (1/c— c) 1— c2 .
If Ip(z) E 9N, then, as required, 1 ____ e2
r=
2 log 1/c •
The function 1 — C2 2 log 1/c
increases monotonically from 0 to 1 as c increases from 0 to 1 and so as Œ decreases from it to O. Thus the equation
r=
sin2 (a/2) 2 log sec(a/2)
has exactly one solution in the range 0 < cc < it.
143
Examples
We denote by 93 the class of functions regular in 1z1 < 1 and such that
/20 = sup(1 —1z1 2 )1f(z)1 < oo.
(4.22)
Izi 0),
where p is a positive number. In what follows we shall call such functions areally mean p-valent (a.m.p-valent). We shall consider in the first part of this chapter some consequences of the more restrictive hypothesis
p(R) p (R > 0). (5.1) Functions satisfying this condition were introduced by Biernacki [1946] and are generally called circumferentially mean p-valent (c.m.p-valent). We first prove some sharp inequalities restricting the growth of functions c.m.p-valent in 1z1 < 1 and such that either f (z) 0, or f (z) has a zero of order p at z = 0, where p is a positive integer, basing ourselves on Theorems 4.13 and 4.17 of the last chapter. In the second part of the chapter we prove some regularity theorems for a.m.p-valent functions in 1z1 < 1. We proved in Theorem 2.10 that if 00 f (z) = V an ? is such a function, then a = lim (1 — r)2P M(r, f) r — ■ 1—
144
5.1 Functions without zeros
145
exists. We shall prove that, if further p> j4 , then also
a
iim lanl n2P -1
A generalization to positive powers of mean p—valent functions and funcak,+,,z k n' will also be proved. As an application tions of the type z
dn=o
—
of these results, we shall prove that if 1(z) = z + a2 z 2 + • •
is a.m. 1—valent in 1z1 < 1, then la rd
5.1 Functions without zeros
n for n > no(f) (cf. Example 2.12).
In this section we prove
Theorem 5.1 Suppose that f(z) = ao + alz + • • is c.m.p—valent and f(z) 0 in 1z1 < 1. Then a1 ^4pao.
Further, unless f(z) = ao [(1+ ze ie )/(1— ze i9 )] 2P for a real 0, we have for p (0 < p 1, and that 0(z) = [f(z)]' possesses a power series expansion (5.4) in an annulus 1-26 < izi 20 .
(5.6)
This is possible since p/1, > 1. Then Schwarz's inequality gives 7.1-
zir
f ir 141(pe)Ids9 —7r
t
t
ir
< ( 217r f l ope/6 )1 2 10(peiordo) 2
(
—n
1 r
2n
L
I 4)(pe w )1 140) 7
,
_i
1( 1
/ 21r f
/0 2
ie (2—tr-2
1
2
A de) 2 X (
1 f 2n If(Pe i° )1 14 d0) . zno Since oc = 0 we can, given E > 0, find ro (E) < 1 such that A — 27r 0
If (pe )1 If(pe )1
M(r, f) .- E(1 — 0 -219 (ro () < r < 1).
(5.7)
Since f(z) is mean p-valent and (5.6) holds, we may apply Lemma 3.1 f
Here and subsequently F(x) denotes the Gamma function. Relevant properties will be found for instance in Titchmarsh [1939] pp. 55-8. We need assume only that f(z)/z' remains one—valued for some real y (cf. Example 2.8).
Circumferentially mean p—valent functions
152
with (2 — 02. instead of 2. Hence if r0 (E) < r < 1 there exists p such that 2r — 1 < p < r and 1 1 2m 4 ,,E ;.(2—t) (pei0)121fi pei0)1(2-0).-2de
02
frro M(X'f)4t x
d
x
P n 2 + 1] fr e t (1 — x)-2pAt dx — [t) 2 X ro
e is chosen so small that pe t [(2t)2 +1] < 1 , (2p4 t —1)
and then r is chosen sufficiently near 1. With this choice of ro, E, r, p we deduce that f2m
27r 0
1(1)'(pei° )Id0
A(p, >it ) e ).(2—t) (1
r ) ;11-213.1t-2p1(2-0-1]
= A( p, ))( 2_t) (1 _ 0-2p, where p is some number such that 2r —1 < p < r. We choose r = 1-1/n and apply (5.5). Then if n is sufficiently large
Rn + ti)bn
-(1 —
2 —" 1 / 2' 0 141(pei° )1(10 n
A (A /1, )E ).(2t)n2P).. Since E may be chosen as small as we please, we deduce
bn = o(n2P;' -1 ) (n —+ +0r4, and this proves Theorem 5.5 when a = 0.
5.4 The case a > 0: the minor arc When a > 0 it follows from Theorems 2.10 and 2.7 that f (z) has a unique radius of greatest growth arg z = Oo and that If (re )1 is relatively small except when 0 is near Oo.
5.4 The case a > 0: the minor arc
153
We shall deduce Theorem 5.5 from formula (5.5) in this case by obtaining an asymptotic expansion for O'(peie ) on a major arc {0 :10 — 00 1 < K(1 — p)}, where K is a large positive constant, and showing that the complementary minor arc y = {O : K(1 — p) 0 and that 00 satisfies (2.59). Then given ri > 0, we can choose K > 0 and a positive integer no with the following property. If n > no, there exists p in the range 1-1/n < p
1, which is possible by (5.6). Then
f
(1
.6,
If (pel )1 -, dO
TE
x —Ât(2)- E) ax
JK(1—p)
K 1—m2p—E) ( 1 _ 01 -)1(2p -e) -E).t n[2t(2p — v) — 1] K 1—)1(2p—e) n[Àt(2p — 8) — 1]
We note that when 2, t, C, ri and chosen so large that
224pc )-(2-0 2(2
—
E
(2n)2t-1
(5.10)
have been fixed, K may then be
K 1--i.t(2p—E)
t) n[242p — E) — 1]
24P).
With this choice of K Lemma 5.3 follows from (5.8), (5.9) and (5.10).
5.5 The major arc Our next aim is to find an asymptotic formula for f (z) , f'(z) and 0(z) on the major arc
F
Fn : 16/
001
K ( 1 — 14
(5.11)
of 1z1 = p. Thus F is complementary to 7. For this purpose we use the asymptotic formulae (2.59) and (2.60) of Theorem 2.10. We define rn = 1 — 1/n. Let p = p„ be the number of Lemma 5.3 and write
= pn ei6), n > 1 = (1 — Pn) 2P f (zn),
(5.12)
and fn(z) =
(1 — ze — i°0) 2P •
(5.13)
We deduce from (2.59), (2.60) that
ccn
n-2P f (rn e'°°)
(5.14)
and kLnI --+ a as n
Using (2.60) we prove
co.
(5.15)
5.6 Proof of Theorem 5.5
Lemma 5.4
155
We have
f (z)/ f n (z) —*
1,
(5.16)
t(z)/f(z)
1
(5.17)
--+
uniformly as n —> cc, while z G F. We write z = pn ei° and recall that by (2.60) of Theorem 2.10 we have
2p
f'(z) f (z)
o(1)
ei°0 — z
uniformly for z on Fn as n —> co. Integrating along F, from z = zn to pn ew , we obtain
log f (p
n eo)
= 2p log (
f (zn)
ewo — z n ei°0 — pnew +
2p log (
lp nePon 00) —
+ o(1 ),
-
1 — pn 1/(2n) and the length of Fn is at most since lei00 — zl 2K(1 — p n ) 2K /n, by (5.11). Thus pne 0-00) 1 2P
f (pneio)1
f (zn)(1 — Pn) 2P
Writing z = Ne i° and recalling (5.12) and (5.13) we obtain (5.16). Next we deduce from (5.13) and (2.60) that f'(z) f n (z) f (z) f(z)
=
f'(z) e l°0 — z 1 f (z) 2p —*
as z ---> ew° on the union of the arcs Fn . Multiplying this by (5.16) we obtain (5.17). This proves Lemma 5.4. An alternative approach generalising the method of Lemmas 1.2 and 1.3 is provided for c.m.pvalent functions by Theorem 5.1 (see Example 5.5).
5.6 Proof of Theorem 5.5 We now conclude the proof of Theorem 5.5 when a > 0, by proving the stronger
Theorem 5.6
Suppose that 4)(z) = z
E +cc
bn zn satisfies the hypothe-
ses of Theorem 5.5, that a > 0 and that 00 satisfies (2.59). Then
nbn
(I) [(1
-'1F 1 )(e2i °°p) ) e—itn+11)°° (n —> +co).
Circumferentially mean p-valent functions
156
We have by (5.14), (5.15)
Ø
f [ (1 -
[( 1 —1 ) el l
oc;-n2Pj- (n —> )
Thus Theorem 5.5 follows at once from Theorem 5.6. The latter result is significantly stronger, since it gives information about arg bn as well as Ibnl. We suppose, as we may without loss of generality, that 00 = 0, since otherwise we may consider f (zewci), 4)(ze 10°) instead of f (z), 0(z). Write
(1 —
E
=
CnZ .
n=0
Then 2p2(2p2 + 1) ...(2p/1+ n — 1) Cn
1.2 ... n F(n 2p/i) (n F(n 1)F(2.1p) F(2/3),.)
-Foo).
(5.18)
We also set O n (z) = f n (z) A , so that On(Z)
=
E cni zm, m=0
and hence
p 1—n 2.7r
1 44(z)dz ncn an = 27riZn 1 1=P
7E
otn (pe i0) e—i(n-1)0 do (0 < < 1) p
.
Finally we recall (5.5): f-Fiz
(n + P)bn =
2n ,
4)'(pei6 )e-i( n+P-1)e d0 (1-26
< p 0 and choose K so large that for n> no we such that can find p = pn in the range 1 — < p 14 , since for p < 1 we cannot improve on Theorem 3.5. We also suppose that either
(i) f(z) is regular and areally mean p—valent in A : Izi < 1, in which case n goes through the integers from 0 to oo in (6.1), or (ii) if p is not an integer we also allow the possibility, as in Theorem 2.11, that f (z)/zP is regular in A and that f(z) is mean p—valent in A, cut along a radius z = pe, 0 < p < 1. In this case n = m + p in (6.1) where m goes through the integers from 0 to co. With these hypotheses we shall obtain estimates for
I lan+i I — lanll. We note that if f(z) is a pth power of a Koebe function
f(z) = (1 where p
zP _) 2P'
1 and n = p + m, we have an+1 — an
=
F(2p + m) F(2p — 1)F(m + 2) m 2p-2
ri2 P-2
F(2p — 1)
F(2p — 1)'
while if
f(z) =
zP (1 — z 2 )P' 165
as
n
166
Differences of successive coefficients
and n = p +m, where m is even, then an+i = 0, and F(p + ,112) an = lan+i — anl =
nP-1
2P-1 F(P) .
F(P)F( 1 +
This suggests the conjecture that
I lan+i
0(n 2P-2 ), p
=
—
p 6,
n
p—n < (1 — —3 ) n < (1 — 1 ) 6 = 64 and n— 1 2 n
. 6 5'
We integrate (6.11) from p= 1 — 3/n to p= 1 — 2/n. This yields 27r f 1-2/n 32n I pe i° — ZIllf (pe 1e )lpdpd0 dO 1-3/n (n — l)n /0
liki an_i — e n l -.
40
1 -
(6.24)
4
and 2n
10
(1 —
r)rdrf G(if(res(9 )1)d0
3 (6.25) p> _ . 4 We apply Lemmas 6.1, 6.2 with v(z) = G(R) and use (6.17). Suppose first that M < 1. Then < c2 min
{M (4P -3)/(2 P) , (1 _ p)3-4p}
G(R)
1. In this case we choose a = b = 1 and obtain in Fk
(1— r)lz i — z1 2 < C0nMk+311(2P) M i
2p)
Also since f is mean p—valent and III < Mk in
conw 2h323k1(20 . Fk
we have
(6.52)
(z)1 2 rdrdt9 < TcpMi = 47.cpM2-2k .
f fFk Thus 2/p 231020 7.cpmi2 -2k 1t(z) 1 2 1 z — z11 2 ( 1— r)rdrd0 0, k is a positive integer and we assume as in (ii) of the introduction that, if p is not an integer, f(z)/zP is regular in A and that f(z) is mean p-valent in A cut along a radius z = rel° , 0 < r < 1. We make a substitution C = Z k and note that F(Ç) = f(C 1 /k )= CP/k (1+ ap+kC + ap-E2k 2 + • • .) is mean (p/k)-valent in ICI < 1 in the above sense. (cf. Example 2.8, or Section 5.8.) In fact a sector 2m S : 0 1 < arg z < 0 1 + — 0 < Izi < 1 k ' corresponds (1,1) conformally to S': kO i < argC < kO i + 2m.
Since F(Ceii) = 0F(ç) we deduce that p(R, S', F) = p(R,S, f) is independent of 0 1 . On applying this conclusion in turn with 0 1 + 27ry/k instead of 0 1 , where y = 0, 1, 2, ...,k — 1, we deduce that p(R, S', F) = p(R, A, f)/ k.
Thus, since f is mean p-valent,
L
R
p(t, S' , F)tdt = k1 f0 RP(t„ntdt S _. kP R 2, 0 < R < oo
so that F is mean (p/k)-valent. We may now apply Theorem 6.2 to F( ) and deduce Lucas' [1969] Theorem 6.3
If f(z) is mean p-valent in IzI < 1 with the expansion (6.58)
then lap+nk —
ap-- -F(n-1)kl
0 and v(r, 0) < M we have 2n
2TcM =
2n
Mu(r,O)d0 f u(r,O)v(r,O)d0
io
Jo
GC)
.
n E 1n(xn2 ± yn2 ),.2n . n=1
Letting r tend to 1 we obtain (6.66). We deduce Lemma 6.7 If 4)(z) is as in Lemma 6.6 and n > 0, there exists C, such that ICI = 1 and
190
Differences
We choose
Ak =
of successive coefficients
lik for k < n,
Ak =
0 for k > n in Lemma 6.6 so that n
1
E 1k-ckz k ,
tp(z) =
k=1
sup 9itp(z). Izi---1
M= Then, by Lemma 6.6,
n
n
n
l ei
0,
(6.75)
193
6.10 The theorems of Dawei Shen
where 0 < 00 < 2n. Then there is a constant Ko, independent of z = pe such that 1F(z)1
K2 = IF(POe ie° )1, there exists pi, such that po
pi < 1 and
If(Pie i°°)1 = R.
(6.78)
Suppose that 1F(z)1= R. If R < K2, (6.76) holds. If R > K2 we choose p i to satisfy (6.78) and apply Theorem 6.1 with z i = z, instead of z1,z2,1F(zi)1=1F(z)1= R, and a = b = 1. This yields 2
R < We have, if 10
—
(6.79)
(1 —p i )(1—p)lz _z,1 2
Ool
—
= Pi + p2 — 2p pi cos(0 — 00) (P1 cos(9 — 00) — )0 )2 + pi sin2 (0 — 0) 4 p i(0 — 00) 2 ' n2
while if
pi. Thus in all cases lz zi 1 2
_2 (0 P1
2 —0
2 (0 —
Po
2
00 2
772
'
Also by (6.77) we have 1 —
R
)
1IP
Thus (6.79) yields
R i1 P < (
1— p)(0 — 002 '
which is (6.76). We can now prove Theorem 6.6. We start with (6.11), applied to F(z) = f(z1), with n+1 instead of n, so that z1 = {(2n — 1)/(2n + 1)}ei°°,
194
Differences of successive coefficients
1 — 3/ (n
p K /
dt9
— 2n+6
This yields
1Pel" — zillr(z)lpdpde
F'(z) 2 F(z) pdpd0) (f lz — zi1 2 1F(z)1 2 )
f
1, there exists R such that R1 < R < kR i and l(R) < n(2/ logk)1. In particular, there exists a sequence Rn , decreasing to zero as n ----o co, such that l(Rn ) ---0 0 as n —0 co. We consider the mapping z = 4)- '(w) of D, onto D i and put 1P(w)
= 0-1 (w) — zo.
7.1 Boundary behaviour in conformal mapping
199
Then the level curves YR in D2 corresponding to lz — zo! = R are the level curves kp(w)! = R. Let l(R) be their total length. Then since ip(w) is it. This gives univalent, we may apply Theorem 2.1 with p(R) ._-. 1, A f: Ri i(R)2dR f ' l(R) 2 dR < < 2 TC A _ < 2 Tr 2 . R — 0 Rp(R) — If 1 is the lower bound of l(R) in R1 < R < kRi , we deduce /2 log k
2n2,
/
0, f (z) is univalent in 1z1 < 1 and If (z)1 < 1. Such a function f (z) will be called a transformation. The transformation w = f (z) maps 1z1 < 1 onto a domain D in 114,1 < 1. We denote by S = Si. the set of all points of 114;1 < 1 not in D, and by d = df the diameter of Sf. We ignore the trivial case when Sf is null and f (z) = z. We shall say that two points z, w on 1z1 = 1 and the frontier of D respectively correspond by the transformation w = f (z), if there exists a sequence z n , such that Izlil < 1 (n
1)
and z n -- z, f (z n ) -- w (n -- ce). Let B = B1 be the set of all points oflz 1 = 1, which correspond to points of S. We note that points of lz 1 = 1 not in B can correspond only to points on 1w1 = 1. We write 6 = 6f for the diameter of B. Our aim is to study the limiting behaviour of transformations when 6 or d is small. In this case D approximates to lw 1 < 1 and f(z) approximates to z. Our first aim is to show that, if either of 6 or d is small, then so is the other, and in this case fl is nearly equal to 1.
If f(z) is a transformation given by (7.4), then we have with the above notation 1 — d < )6' < 1. Lemma 7.3
The inequality fl < 1 follows from Schwarz's Lemma. To prove 1 — d, we may suppose that d < 1. It follows from Lemma 5.2 that $
201
7.2 Transformations
if S meets 114, 1 = r for some r < 1, then S meets lwl = p for r < p < 1. Hence S has at least one limit point e'0 on jw1 = 1. Thus S lies entirely in lw — e 0 1 < d, and so in 114/1 > 1 — d. Thus D contains the disc 1w1 < 1 — d. Hence the inverse function z = f -1 (w) maps 1142 < 1 — d onto a subdomain of 1z1 d that arc, cR say, of I w — wol = R which lies in 1w1 < 1, lies also in D. Let yR be the image of cR by z = f-1 (w), and let 1(R) be the length of YR. Then by Lemma 7.1 with R1 = d, k = we can choose Ro satisfying
1 _1 n (-1 log —1 ) 2 n (-1 log —2 ) 2 2(5, z < 1. We may without loss of generality assume that 00 = 0. Then we have by Poisson's formula for 1z1< r 26, 101 < 6, and r is sufficiently near to 1, then 26
re4 z !TN' — z
1—z
Hence we deduce from (7.5), as r
ip(z)1+ z 1 1 6
56
1+z
z1 2.
1,
u(re) `4) (14)
O. Clearly f (z)/ 16 G S. We have further Lemma 7.7 The functions f(z)/fl, where f(z) is constructed as in (7.12), form a dense subclass S i of S. Suppose that f(z) E S and 0 < p < 1. Then f(pz) maps IzI < 1 onto the interior of an analytic curve, namely, the image of Izi = p by f(z). Also Pif (pz) E S 5 if 1, uniformly for Iz I r, when p (pz) -- f (z) as p
0 < r < 1. It is thus sufficient to show that the functions if (pz) can be approximated by functions in S i . Next if M is large 1 1 w = îp(z) = — f(pz)= — z + • • • M PM
maps IzI < 1 onto the interior D of a closed analytic curve F lying entirely in I w I < 1. Let P = rei° be a point of largest modulus on F. We obtain the arc PP,, of F by going along the curve F in an anticlockwise sense from P until the whole of F apart from an arc of diameter 1/n has been described. Then ooPPn is a slit F. Let it be the complement of Fn and let
f n(z) =f3 (z + a2z 2 + •
)
map Iz < 1 onto An . To prove Lemma 7.7 it remains to show that, if t < 1, f()
.1P(
)
,
as n
CO
,
uniformly for ICI
t.
(7.13)
206
The ',Owner theory
For then fn (z)/ fi n approximates p-1 f(pz) for a fixed p, which in turn approximates f(z). We consider gn(z) = fn-1 {tr(z)}.
We verify that gn (0) = 0, gn'(0) > 0, so that g(z) is a transformation. Let Sg, be defined as in §7.2. Then Sg„ consists of all those points in 1(1 < 1 which correspond to points outside D by w = f n(C). Choose now 6 so small that the circle of centre P and radius R meets D in a single arc TR for 0 < R < 6, lies in 1w1 < 1, and that the origin w = 0 lies outside this circle. Then if 1/n < R, y R corresponds to a single arc cR in 1(1 < 1 by = f n-1 (w) and all points of Sgn are separated from C = 0 by CR. By Lemma 7.1 we may choose R so that the length l(R) of cR satisfies 1 l(R) n log(n6)] 2 , and so by Lemma 7.2 the diameter d(g) of Sg„ satisfies 1
d(g) .. n [i log(n6)] 2
1
It now follows from Lemma 7.6 that
gn (z) —> z (n —> co), uniformly in 1z1 < 1. We note that w = fn (C) maps ICI< 1 — d(g) into D and so into 1w1 < 1. Given 0 < t < 1, we now choose p = 4( 1 ± t) and suppose that n > no so that d(g) < (1 — p) = -41 ( 1 - t). Then for 1C1 p we have by Cauchy's inequality 1 2 If/n(01 < 1 — d(gn ) — il < 1 — p . Hence if 1C11
P, 1C21
p, n > no we obtain
1
C2
If n(i) — fn(2)1 = Suppose now that 11
f)(1C
n1 say, we have gn(01 .. p. Now
IfnG) — 001= Ifn(C) — fn{gn( C)}l
2 1 — p IC — gn(C)1 —> 0 , as n
This is (7.13) and the proof of Lemma 7.7 is complete.
207
7.5 Continuity properties
7.5 Continuity properties Following L6wner we now investigate the class el and show that the functions f (z) in (7.12) can be obtained by a series of successive infinitesimal transformations from w = z. Let y be a slit given by w = a(t) (0 < t < co). We denote by ye t, the arc t' t t" of y and by y, the arc y. Let G(t) consist of the complement of y t . As t increases from 0 to co, G(t) expands from G = G(0) to the whole plane. We denote by
w = gt(z) = fi(t) (z a2(t)z 2
• -)
(Mt) > 0),
the function which maps 1z1 < 1 onto G(t) and proceed to show that gt (z) varies continuously with t, as t increases, from go(z) = f (z). We have first Lemma 7.8 If w = g(z) is defined as above, then the inverse function z = g 1 (w) remains continuous at w = a(t). Thus as w = g(z) a(t) in any manner from G(t), z approaches a point A(t) such that 12(01 = 1. Choose 6 so small that the circle 1w —a(t)1 = R meets yt in exactly one point for 0 < R < 6. The choice is possible since yt has a continuous tangent at a(t). Then for 0 < R < 6 the circle 1w — a(t)1 = R lies in G(t) except for a single point. The image of this circle is an arc cR lying, except for end—points, in 1z1 < 1. By Lemma 7.1 we can then choose a sequence Rn such that l(Rn ) 0 and Rri 0, where l(R n ) is the length of cRn . If Rn < WO, then the disc 1w — a(t)1 < Rn cut along the arc yt corresponds to one of the domains into which cR, divides 1z1 < 1, namely, that one, An , which does not contain z = 0. By Lemma 7.2 the diameter of A, is not greater than l(Rn ) and so tends to zero as n ao. Since An c An__i when R, 0, where eig5 is a point of B ee, so that ekl) —> 2(t). Now Lemma 7.10 follows. We can now prove part (b) of our fundamental theorem.
Theorem 7.2 If f(z) G S i so that f(z) is given by (7.12) with /3 = 1, if t < oo, then further g(z) is defined by (7.15) and f(z,t) = [f(z)J, 0 w = f(z,t) satisfies the differential equation (7.1) and f(z,0) = z. Also g(z,t) = et f (z, t) satisfies (7.2) and (7.3). The function w = f(z) maps lz < 1 onto the w—plane cut along the slit y. Also = e(w) maps 1z1 < 1, the plane cut along the smaller slit Yt onto 11 < 1. Thus = f (z, t) maps 1z1 < 1 (1,1) conformally onto a subset of (1 < 1. Using also (7.15) we see that et f (z , t) E e so that (7.2) holds. Also, when t is large, y t is a ray given by (7.11) and w = gt (z) =
4rz
where 4r = e t. Suppose that 1z1 p < 1, so that If (z)! < K, where K is a constant. As t —> oo we see that z —> 0, w 4rz uniformly for 114, 1 < K. Thus f (z,t) = gï l {f (z)}
f (z)
4r
= e—t f (z)
as t —> co, uniformly for 1z1 p. This proves (7.3). Next write f (z, t') instead of z in Lemma 7.10. This is permissible since < 1 for 1z1 0, while t' or t" remains fixed. If t = t' is fixed this gives the required result for the right derivative. If t" remains fixed the required result for the left derivative also follows. In fact by Lemma 7.9 and since K(t) = 1/2(t) K(t')
> K(t") (t'
-
> t"),
-
while t" is fixed, and by (7.14) h(z, t', t") - z -> 0 (t' -> t"),
uniformly in 1z1 < 1. Writing f (z, t') instead of z we obtain f (z , t") - f (z, t') --÷ 0 (t' -+ t"),
uniformly in 1z1 < 1, so that (7.16) gives, also if t = t" is fixed, 1 + K(t)f (z , t) f (z, t) - f (z , t') —+ f (z, t) 1 - K(t)f (z ,t) t - t' as t' -> t from below. This proves (7.1) and completes the proof of Theorem 7.2.
7.7 Completion of proof of Theorem 7.1 In order to complete the proof of Theorem 7.1 it remains to show that given K(t), as in Theorem 7.1, there exists a unique solution w = f (z, t) of the differential equation (7.1), such that f (z, 0) = z, and further that et f (z, t) E It is convenient to put
a.
1 W = U + iV = log (- ) . w Then (7.1) becomes
1 + K(t)e -w
3W at
1 - K(t)e - w
(7.17)
= 0(t, W),
say. We note that 90(t, W) > 0 for 91W > O. Also if U > 6 > 0,
12K(t)e-w I
11
-
140e -1'1'1 2
6, 91W2 > 6,
a oaw IdW
10 (t ,11/2)— (t, W 1 ) 1 fww:
1 KIW2 — W11.
(7.19)
Thus 0(t, W) is Lip in W in the half—plane U > 6 uniformly with respect to t. We now define a sequence of functions Fn (t, co) (n > 0) as follows. We suppose that 91co > 6 > 0 and set
Fo(t, co) co, } Fn (t, co) = co + fc; 4[T, Fn_i(T, (0)1d -c (0 < t < oo, n _. 1).
(7.20)
Since K(t) is measurable so is cf)(t,w) for fixed co. Also 0(t, co) is bounded, so that the integral exists. We note first that 9iFn (t, co) increases with t for fixed co and so remains greater than 6. For if this is true for n, then 910[T,Fn (r, co)]
>0
and the result remains true for n + 1. Next we have K n tn in
IFn(t, CO) — Fry-1(t5 (01
n!
tU
< t
—
< 00,
n > 1).
In fact if n = 1 we have from (7.18) t
t
Fi (t,co) — Fo(t, co) =
fo Kch = Kt.
fo
Also if our result is true for n, then (7.19) gives t I Fn-Fi(t, W) — Fn(t, (0 )1 = f {0[T, Fn (T, 01 — 0[T, Fn_ier, wnIch
o
t
t
Kt j K n+1 tn+1
1Fn(T,(0) — Fn—i(T, (0)IdT K f o at = n! (n + 1)! . o Thus the result is true for n + 1. It follows that the sequence Fn (t, co) converges uniformly for 9ico > 6 and 0 < t < to to a limit function F (t, co). Taking the limit in (7.20) we deduce t (7.21) F (t, co) = co + f 0[-c, F(T, co)]ch .
Kf
JO
Thus 9iF (t, co) _. 6, and so, by (7.18), the integrand in (7.21) is bounded by K. Hence F (t, co) is absolutely continuous in any finite interval
7.7 Completion of proof of Theorem 7.1
213
0 < t < to, and satisfies (7.17) almost everywhere, and conversely an absolutely continuous function W(t) = F(t,co), which satisfies (7.17) almost everywhere and W(0) = co, also satisfies (7.21). Next we prove that F(t,co) is analytic in co. Since Fn (t,co) converges to F (t, co) uniformly for fixed t and 91co > 6, it is enough to show that Fn (t,co) is analytic in co for n = 0,1,2,... We prove by induction on n that Fn (t,co) is analytic and further that, for 91w ' > 6, 9ico2 > 6, t > 0, we have
IFn (t,w2) — Fn(t,(01)I
eK `10)2 — 0)11.
These results are clearly true for n = 0 by (7.20). Suppose that they hold for n. Then (7.19) and (7.20) yield
f' H_1(t,c02) — Fn+1(t,(0 1)1
1 (1)2 — a) 11
1+
f
KeKtcl-c = eKro2 — (D i .
Also {(¢[T,Fn(T,(-0 2)] (P[T,Fn(t,(0 1)]}/( 0)2 — coi) is bounded for 0 < t < t and tends to
1 1- 0[T, Fn (T,04]1 = L ""
[ aFn n( : t C°
1 [ a w, W) L
Fn(T,(01)
as co2 —> coi through any sequence. Thus by Lebesgue's dominated convergence theorem (Titchmarsh [1939, p. 345]) Fn+i(t, W2) — Fn+1(t, (1) 1) = 1 o t [t Fn(' + I (0 2 (0 1 1
[T,
— [t F r 0)2 — (0 1
Fn (T, (0)]
0
ut
ch (0,
as co2 —> col through any sequence and so generally. Thus Fn+i(t,co) is analytic and the inductive step is proved. Suppose next that W(t) is any absolutely continuous solution of (7.17) which satisfies 91147 (0) > 6 and so 9l14/(t) > 6 (0 < t < to ) and further that
W(t i ) = F(ti ,co) for some pair (t i , co), such that 0 < t 1 < to and Who > 6. W(t) F(t,co) (0 < t < to ). In fact write M = sup 1W(t) — F(t, COI 0 -1 (0
This gives
(1z1= r).
(z)}
The inequality (1.6) of Chapter 1 gives for f (z)
r(2r — 4)
(1z1= r),
so that our condition is satisfied if 2r2 — 4r > r 2 — 1, i.e.
r2 — 4r + 1 0, 0
r
2—
0
2g).
227
Examples
Thus f (z) in S maps Izi = r onto a convex curve for 0 < r < 2 — .0. On the other hand if f (z) = z(1 — z) 2 , then
Z
f"(z) 2z2 + 4z = f'(z) 1—z 2 '
and this is real and less than —1 for —1 < r < V3-2. Thus this function does not map Izi = r onto a convex curve for r > 2 — V3. The quantity r, = 2 — V3 is called the radius of convexity.t We may ask similarly for the radius of the largest circle 1z1 = r such that the image y(r) of Izi = r by f (z) always bounds a starshaped domain with respect to w = O. The condition for this was seen in Chapter 1, (1.15) to be that
} 0, f(z) 9q z f(z)
i.e.
TE
f'(z)}
arg {z f(z)
(1Z1 = r).
If we write
f 0(z) =
(41 + z 1 + fo z
)
— f (zo)
(1— I zo1 2 ).r(zo )
'
(7.41)
then 0(z) E S if f(z) E S. On applying the inequality (7.37) to 4)(z) at z = —zo, we obtain
arg
(
f( zo) zo f' (zo))
)
...
log
1 + lzol 1 — Izol'
(7.42)
and this is sharp. Thus the radius of starshapedness, rs, being the radius of the largest circle whose interior is always mapped onto a starshaped domain with respect to w = 0 by f (z) E S, is given by
n 1 + rs n — = log I-, = tanh — = 0.65... 2 4 Examples
7.5
If f(z)
E
S and (/)(z) is defined by (7.41) show that
zo f(zo)(1 — 1z01) 2. Deduce that (7.42) is sharp and verify (7.43). —
t Gronwall [1916] * Grunsky [1933]
z 0 0'( —zo) Sb( — zo)
(7.43)
228
The L5wner theory
7.12 The argument of f'(z) result of Golusin [1936].
Theorem 7.7
As a final application we prove the following
Suppose that f(z) E S. Then we have the sharp inequalities
{ I arg f(z)I < 4 sin-1 Izl I arg f(z)I < n + log
(1z1 --v -) t i 1) .
1z1 2 1 —Iz1 2
,/2) IZCII2
ilf(zo,to)1 0 1 — "(' )
if If(zo, to)I < and so for all large to . Thus the upper bounds of Theorem 7.7 may be approached as closely as we please and so are sharp.
7.13 Conclusion The foregoing theorems represent some of the principal successes achieved by L6wner and his successors by means of Theorem 7.1. In the next chapter we shall use the technique to prove de Branges' Theorem and some of its consequences. At this point Schiffer's variational method and Jenkins' theory of modules t should be mentioned. These methods can be used to prove the results of this chapter and some others and in particular to give more information about the extremal functions. However Liiwner Theory has so far proved to be an essential ingredient of the proofs of de Branges' Theorem. Schiffer [1943], see also Duren [1983, p. 318 et seq.] Jenkins [1958].
8 De Branges' Theorem
8.0 Introduction In this chapter we prove de Branges' Theorem [1985], conjectured by Bieberbach [1916] that, if x, anz n E a.-,, (8.1) f(z)=z+ we have Ian I < n, n= 2, 3, ... with equality only for the Koebe functions f(z) = z(1— zeie) -2 . This result had previously been proved for n = 4 by Garabedian and Schiffer [1955], for n = 5 by Pederson and Schiffer [1972] and for n= 6 by Pederson [1968] and Ozawa [1969]. (For a more detailed history see Duren [1983, p. 69].) De Branges proved his theorem by first establishing a conjecture of Milin [1971], which Milin had shown to imply Bieberbach's conjecture. Suppose that
log
f(z)
k CkZ ,
z
(8.2)
then Milin conjectured that n
E ( _4k _ 1 ( I
c k 1 2 ) ( n—k
+1) 0, n= 1, 2, ...
(8.3)
k=1
The proof of de Branges has been simplified successively by Milin [1984] and Emelyanov, by Fitzgerald and Pommerenke [1985] and Weinstein [1991]. Inevitably these simpler proofs however miss the operator theory basis of de Branges' subtle ideas. All the proofs rely on Li5wner Theory, and a positivity result for the coefficients of certain special functions. The earlier proofs used an inequality of Askey and Gasper [1976] concerning Jacobi polynomials. Weinstein's proof, which we shall follow here, uses instead the addition formula for Legendre polynomials which
230
8.1 Legendre polynomials
231
goes back to Legendre himself and seems simpler to establish. However Wilf [1993] has now shown rather surprisingly that the two results are equivalent. We start off by proving Legendre's formula, then prove Milin's conjecture, Milin's inequalities and de Branges' Theorem, which in fact generalises to give sharp bounds for the coefficients of (f(z)/z);-, when ), > 1. The analogous result fails for 0 < < 1. A number of further generalisations and consequences, including proofs of conjectures, of Robertson [1936] and Rogosinski [1943], will be given at the end of the chapter. De Branges' Theorem, both the result itself and the subtlety of its proof, represents a milestone of twentieth century analysis.
8.1 Legendre polynomials
The Legendre polynomials P(z) are defined
by the expansion 0,
1
=1
+ E hnPn (z),
(8.4)
(1 — 2zh h 2 ) 2 valid when 1111 is sufficiently small depending on z. We follow the account given in Whittaker and Watson [1946, Chapter 15] and start with the following simple lemma, that is easily proved by means of the calculus of residues.
Lemma 8.1 Suppose that A, B, C are complex numbers, such that A 0, 0, that F is a circle containing in its interior A = B 2 — 4AC 0)/(2A) but not t2 ( —B —/A)/(2A), and that F is described t1 = in the anticlockwise sense. Then dt fr At 2 Bt C
27ri \./A •
(8.5)
We shall also need the associated Legendre functions pnk ( z ) = (z 2
1 ) ,;k
d k
Pn (z), 0
We suppose for the time being that 1 < z < c and positive. We choose for F the circle t = Z
(Z 2 — 1) 2 e iu ,
— TE
1. Thus } pnk( z ) = (n + 1). • • (n k) {z + (z 2 — 1) COS 0 n COS k0c10 (8.7) 2m
and k
Pn (z) =
X
f
n(n — 1) • • • (n — k + 1)(-1)k 2n —n-1
+ (z2 — 1) 2 COS 0)
COS
Ode.
(8.8)
We can now prove Legendre's addition theorem. Lemma 8.2 Suppose that x, y, w are complex numbers and that Z = xy — (x2 — 1)1 (y 2 — 1)1 cos co. Then P1 (Z) = P(x)P(y) + 2E n (-1)k (n k)! P k (x)P k (y)cos kw. (n + k)! n n k=1
8.1 Legendre polynomials
233
For the proof we shall assume that x, y, co are real, x > 1, y > 1, and that the positive square roots are taken in the definitions of Z and Pnk (x), e;(Y). The general result, with a suitable definition of the signs of square roots, is then obtained by analytic continuation. We assume that h is small, and that 4) is real. Then I
{X ± (x2 — 1) 2 COS(0) - o
x
E h"
{y
n=0
i
± (y 2 — 1) 7 cos 0}
r
n+1
1
=
.
(8.9)
y + (y 2 — 1)-1 cos s 4) — h (x + (x2 — 1) 1 cos(co — We integrate both sides of this equation w.r.t. 4) from —7r to n. The integral of the right-hand side takes the form i
=
r J _ir
dck a + bcos 4) +csinck'
where
a = y — hx, b= (y 2 — 1) -1 — h (x 2 — 1) cos co, c = —h (x 2 — 1) I sin co. To evaluate / we write t = eiO , dck = dt I (it), cos ck = 1 (t + t'), sin 4) = (t — t-1 ) /(2i). This yields
I =
dt li _ i At 2 + Bt + C
2ni \IA'
by Lemma 8.1, where A = -1-(c + ib), B = ai, C = 1(—c + ib).
(Since IA I = ICI, the Lemma is applicable.) Also A
= B 2 — 4AC = b 2 ± C 2 — a2 = (y2 _ 1 ) ± h2 (x 2 1 ) 2h (x 2 — 1) / (y 2 — 1) cos co — (y — hx) 2 1
= — {1 ± h 2 — 2hxy + 2h (x 2 — 1) / (y 2 — 1) 2 cos co} = — { 1 + h 2 - 2hZ } . Thus
1 =
+27r .\/ (1 + h2 — 2hZ) •
234
De Branges' Theorem
To check the sign, we put h = 0 and obtain 77r
=
= 27r,
y + (y 2 — 1) cos (1) by (8.8) with k = n = 0. Thus for small h we have 2
=(1+
— 2hZ) —
h"P,(Z) n=0
by (8.4). On comparing coefficients with the integral of the left-hand side of (8.9) we obtain 1n {X + (X 2 —
P(Z)
= 27C fir
2 COS(W — 0)} d4)
{y2 ± (y2 ____ 1) cos
(8.10)
}n-4-1
•
The right-hand side is a polynomial of degree n in cos w and sin co and is an even function of co, as we see on making the substitution 4) = Thus 1 P,(Z)= —A0 ±
2
Ak cos kw. k=1
It remains to evaluate the coefficients Ak. We have Ak =.-
it
—77
1
Pn (Z)cos(kco)dco
i
n
_ircos(kw)dco
27,2
r
x + (x2 —
1) 2 COS(CO
o) y do ,
n+1
•
fy + (y 2 — 1) 1 cos (/)
We integrate first w.r.t. co, setting co = Ø + 0 and using (8.7). We integrate from w = — it to 4) + 7r, i.e. from 0 = —n to n. This is legitimate since the integrand is periodic in co. We also write cos kw = cos Ice cos k4) — sin Ice sin k4) and note that the integral containing sin(k) vanishes since the integrand is odd. Using (8.7), and for the remaining integral in 4) (8.8), we obtain Ak
2/3,` (x)P,;((y)(-1 )k (n+k)(n + k — 1)• • • n(n — 1)• • -(n— k +1) 2(n — k)!(-1)k P(x)Pl(y). (n+k)!
This proves Lemma 8.2.
8.1 Legendre polynomials
235
8.1.1 We complete the section by proving the positivity result which is needed for Weinstein's proof. (In a recent preprint by Ekhad and Zeilberger [1993] the authors produce a short direct argument for this.) Lemma 8.3
Suppose that t > 0, and that w = w(z) is defined by
et w (1 — w) 2
w,(0) = O.
(1 — z) 2 '
(8.11)
Then we have for z1
O.
Now Lemma 8.2 yields
P(Z)
Ak,„(t)coskw, k=0
where Ak, n (t) > O. Next 2
Pn (Z)zn}
1 —2zZ z 2 {r.,
=
Z
m
In=0 k=0Ak, ni COS kw} {Êzn
E n Ai, n COS /W}
n=0
We recall that cos kw cos lw = {cos(k — obtain 1
1 — 2zZ z 2
n=0
1=0
+ cos(k + Ow). Thus we
Bo i COS k=0
.
kco,
(8.13)
236
De Branges' Theorem
where Bk, n (t)
O. Again by (8.11)
1 z + — 2Z
1 —2zZ + z 2
1 et w e— t ( 1 w — 2) — 2Z 1+ w 2 — 2w cos w
2
1 w2 1 w 2 — 2w cos w x etw { ,k cos(kw)} 1 _ w2 + 1 2 k=1
=
=
Ag(t)z" +1 ± 2 k=0
iVi (t)zn +1 cos kw. k=1 n=k
then of zn+ 1 cos kw in this and
Equating coefficients first of z' (8.13) we obtain that
AP(1 = B,, and Afkl =
k>1
and this proves Lemma 8.3. We remark that wt (z) maps izi < 1 onto 114, < 1 cut from —1 to --- "C along the real axis, where
= 2et —1 _ 20e2t — et). Thus wt (z) is regular and Iwt (z)i are valid for lz! < 1.
for 1z1 < 1 and so the series (8.12)
8.2 Proof of Milin's conjecture: preliminary results
In this section we
prove
Theorem 8.1
If f(z)
Ee
and
log
f(z)
GC
(8.2) k=1
then we have for n = 1, 2, ...
4
— kIck1 2 ) (n — k + 1)
O.
(8.3)
k=1
Theorem 8.1 is the key step in de Branges' proof. As we shall see the result also leads to a number of conclusions which go beyond Bieberbach's original conjecture. Weinstein's proof actually shows that equality
8.2 Proof of Milin's conjecture: preliminary results
237
holds in (8.3) only for the Koebe functions fo(z) = z/ (1 — zei 6 ) 2 . He uses a refinement of L6wner Theory which applies to every function in S and not just to a dense subclass. For this we refer the reader to Weinstein [1991] and Uiwner [1923]. To prove Theorem 8.1 it is sufficient by Theorem 7.1 to consider the functions f(z) in the subclass S i of that theorem. Thus
f(z) = lim et f(z,t), where f(z,t) satisfies Uiwner's equation (7.1), K(t) is continuous and k(t)! = 1. We define g ( ) as in Theorem 7.2 by
= f(z).
gt If(z,
(8.14)
Thus go(z) = f(z). Also for large t et z
gt (z) =
(1 — zei0) 2.
We now write OC
h(z, t) = log
gt(z)
ck(t)z k .
zet
(8.15)
k=1
Thus ck (0) = ck,
2 k
and ck(t) = —e 1-4", t > to.
(8.16)
We deduce from (7.1), (8.14), just as in Section 7.8, that (7.23) holds, i.e., if g = gi (0, = f(z,t) we obtain
eg eg r 1 + et
404. 1—
K
g = g1 (z) we have
or writing z instead of
eg äg 1 + KZ —= Z et äz 1 — KZ
(8.17)
Using (8.15) we deduce
0
— h(z,t) = et
0 — h(z, t) ez
=
eat
eg/ez g
1 1 z•
Thus (8.17) yields Oh (1 0h) 1+ Kz 1+=—+ z z ez — KZ et
238
De Branges' Theorem
i.e. Oh = ( 1 + z a h) ( 1 + Kz) — 1. at az ) 1— . KZ We substitute (8.15) in this and equate coefficients in the resulting series. This yields c(t) = kc k 2k k +2
rCr K
k—r
.
(8.18)
We need some crude bounds for ck(t),c(t).
Lemma 8.4
We have for 0 < t < oo and 1 < k < oo,
1001 < 9, 14(01 < 11k 2 . We deduce from (8.15) that z
ag/Oz
=1+
k ck z k
.
Also e'gr (z) E S. Thus (1.4) and Cauchy's inequality yield 1 —k- sup r I z i =r
1(10(1
Oglez
r
(8.26)
and 01 — 0 1+1
0-k+1
0-1c+29
1
k < N —1.
(8.27)
0•
(8.28)
Then E (kIcki2 — —4 ) o-k
We choose nonnegative numbers (xi to aN, multiply (8.3) by —a n , and add for n = 1 to N. We deduce that 2 4 E (klcd — — k
ak
0,
k=1
where ocn (n — k +1).
=
(8.29)
n=k
It remains to show that if the o-k satisfy (8.26) and (8.27) we can solve this system of equations for nonnegative oc n . In fact (8.29) yields ŒN = 0- N
and if k 0. By induction on n we deduce that Ak > 0 and Dk > 0, for k = 1 to n. If O i * 0, we apply the above argument with w 1 (z) = w (ze —i°1 ) and 4)(z) = (ze —i81 ) instead of w(z), 4)(z). Then Ak, 1Dk are unchanged so that equality still holds in (8.33) but now ai (0) > 0 so that the first n coefficients of w i (z) are positive. Thus
Ak = Ake1
=
e k01 ,
k = 1,
n•
This yields (8.34) with ri = e 0' and the proof of Theorem 8.3 is complete.
246
De Branges' Theorem
8.3.1 We can also obtain a corresponding bound for Dk. This is Milin's second inequality [1971, Theorem 2.4, p. 50].
Theorem 8.4
With the hypotheses of Theorem 8.3 we have
113 ,1 ._ dn(A)exp
n
1
dn kw ( k 2 01(1 2 — 22
{ 24:1)
(8.41)
kA
)
Again equality holds if and only if (8.34) holds.
To prove Theorem 8.4 we deduce from (8.35) and (8.33) for n = 1,2,... n
1 IDnI 2 < n2
n
-
vdn_v(il) v—
v=i
1
Ip v 1 2 d,(.1.)
< dn_1(i1-1- 1) n2
n
1 dn_1(A +1)
n _ v (1) exp { v=i
n—v-1(A + 1 ) } .
(8.42) We have, since x < ex -1 n
E l a d 2dn
_ r( 2
,
)
v=1
= /12 4- 1( 2
n
1
+ 1)
{
42,dn—v().) ,1 2 dn_1(2 + 1 ) v=
< 22 dn-1(il + 1 ) ex P { ,1 2 d
}
n
1
1
(A + 1
}
1
. A 2 c/ n- 1 (A, + 1 ) ex P { /1 2 d 1 _1( 2
+ 1)
(
V
22 )
V=1
by (8.39). On combining this with (8.42) and using (8.32) we have IDnI __
n
22) (d„—v-1(/1. + 1 ) Vil
fi
= d1(2)exP 1
n
2dn(2) V=1
(a,, _ 22 ) Av
dn—v( 2 )} •
+
dn—v(1 /12
8.4 Proof of de Branges' Theorem
247
This proves (8.41). Equality holds only if equality holds in (8.42). This yields equality in (8.33) with n — 1 instead of n, and so (8.34) holds with n-1 instead of n. Also equality holds in (8.35) with k = n and this shows that (8.34) holds also for n. This completes the proof of Theorem 8.4.
8.4 Proof of de Branges' Theorem conjecture and a little more.
Theorem 8.5
We can now prove Bieberbach's
a and that = Ê anwz n—i .
Suppose that f(z) E
( f (z))'
(8.43)
z)
n=1
Then, if A > 1, and n> 2, we have
lan(4 __ d n _1(2.1) =
F(n — 1 + 22) F(n)F(22) •
(8.44)
Equality holds if and only if f(z) is a Koebe function
k8(z) =
z (1 — zei8 ) 2
, where 0 is real.
(8.45)
If 2 = 1, ci„(2) = an , we obtain Bieberbach's conjecture. We also recall from Example 7.3 that (8.44) fails for —1
F(2 + 22) =14 22 + 1 1 - -1d2( 24. 2F(2)
To prove (8.44) we deduce from Theorem 8.1 that (8.3) holds. Thus
= exp
ACkZ k
{:=1
,
}
where n
, 4 E (kIck l- _ —) (n — k + 1) 0, n = 1, 2, ... k k=1 We apply Theorems 8.2, 8.4 with p = 2 2 instead of A, and Ak = ACk, Dk = ak +i(A). In order to apply Theorem 8.2, we need to check that, if
248
De Branges' Theorem
ak = dN—k(P), k = 1,2,..., N +1, p our convention
2, (8.26) and (8.27) are satisfied. By
0-N+1 = d-1(p) = 0. Also o-k > 0 for 1 < k < N, and by (8.32) Crk+1
dN—k-1( 1 )
ak
dN—k(J1 1 )
N—k 1. p+ N—k—1
Thus (8.26) holds. Next N k ak -Ei — ak = ( p+N—k-1
= (1
1 ) uk, ak
p N — k) ak• 1N—k
Thus { Crk+1
20-Ic
ak-1 = ak
1—p
1—p p+N—k-1 1+N—k}
— 1 )(P — 2)ak >0 (N —k p-1)(N —k +1) —
if 1 < k < N. Hence (8.26) and (8.27) are satisfied and we deduce that
E(kck12—
(k2 lAk 1 2
2
4
4A2 ) dN_k(2)1.)
dN—k(P)= -AT;
k=1
O.
k=1
Now (8.41), with
instead of)., yields
laN+I(A)1= This is (8.44). If N particular we have
dN (2).), if N > 1.
1, equality is only possible if (8.34) holds. In =
=la2(1)1= 2.
Now it follows from Theorem 1.1 that f(z) is given by (8.45).
8.5 Some further results We note some consequences of Theorem 8.3, which go beyond Bieberbach's conjecture. Theorem 8.6
If f2(Z)
a21-1Z 2" G S,
then
.a2N-1, 2 < N lail 2 +1a3i 2 +... + II
1.
(8.46)
Equality holds for N > 2 if and only if f(z) = z/(1 — z 2 eie ), for a real O.
8.5 Some further results
249
The result was conjectured by Robertson [1936] and proved by de Branges. We recall from the end of Chapter 5 that, since f 2 (z) is odd and univalent, f (z) = 2 (z )1 2 is univalent. Thus f2 (z)
1 + a3 z + a5 z 2 ± • • • =
(f(z)) 4 Z
z
Hence with the notation of (8.43) we can write (8.46) in the form
E l an( ) 12
(8.47)
N.
k=1
To prove (8.47) we apply Theorem 8.3 with follows from Theorem 8.1 that
k= 1
k 2 lAkI 2 k
)'2 ) )
Ak = ..Ck
041( 1 2
n
4
k=1
1)
k
and A = 1. Then it
(n—k+1)_ O.
Now (8.33) yields
ElDk12
n + 1.
k=0
Equality holds for n > 1 only if la3 1 = 1. In this case
f (z) = z + 2a3 z 2 + • • and since f (z) E S, f (z) = z (1 — zei°) f2(z)
=f
2 by Theorem 1.1. Thus
1
( z 2) 2 =
(1
z 2 ei0) •
This proves Theorem 8.6. Theorem 8.6 implies a conjecture by Rogosinski [1943], as was shown by Robertson [1970]. Theorem 8.7 Suppose that w i (z), w 2 (z) are regular in 1z1 < 1 and satisfy < 1 there, and further w2 (0) = 0. Suppose that f(z) E and that F(z) = wi(z)f { 02(z)} =
Z
n .
n=1
Then lA n 1 < n, n = 1,2,... Equality holds if and only if F(z) = e'Â ko(z), where k 0 (z) is the Koebe function (8.45).
250
De Branges' Theorem
If co i (z) EE 1, then F(z) is said to be subordinate to f(z). This is the case for instance if f (z) maps Izi < 1 onto a simply connected domain D and if F(z) is any function regular in 1z1 < 1 and with values in D. For these functions F(z) the result was conjectured by Rogosinski [1943]. To prove Theorem 8.7 we follow the exposition of Duren [1983, p. 196]. We write
i Z.) 2 (z n
H(z) =1
=
z ± c3z 3
and proceed to show that n l Ai
E 1C2k-112
(8.48)
n