level sets of univalent functions

Comment. Math. Helvetici 56 (1981) 3 6 6 - - 4 0 3 0010-2571/81/003366-38501.50+ 0.20/0 O 1981 Birkhiiuser Verlag, Basel

Leve l sets of univalent funct ions

W. K. HAVMAN and J.-M. G. Wu*

1. Introduction

Let w = f ( z ) be a univalent (one to one analytic) map from za : [ z l < l onto a domain O in the closed complex plane. By a level set of f we mean the preimage of f - l (g] n L) for some straight line or circle L. Piranian and Weitsman asked in [7] if every level set of f has finite length. Here we give an affirmative answer. We shall denote by A0, A1 . . . . positive absolute constants. If E is any set, 6(E) is the d iameter and IEI is the length or one-dimensional Hausdorff measure of E. Then our result is

T H E O R E M 1. I f E is a subset of a level set of a univalent function, then

IEI-< AoS(E) <- 2A0 (1.1)

where A o < 10 35.

A special case of the theorem when O is a Lipschitz domain was proved in [8]. Our argument is rather lengthy and falls naturally into three parts. In the first

part (Sections 2 to 5) we assume that E is the full level set 3'. The components , i.e., maximal connected subsets of 3", will be called level curves and denoted by 3"k- We shall then prove in Section 5,

~ ( 3 " k ) ~ A1 =2.1 x 1016. (1.2)

About three weeks after we obtained this result a proof of (1.2) was obtained independently by Gehring and Jones [1]. It is also worth noting that Jones [4] has constructed a bounded analytic function on A for which every level set I wl = R is ei ther empty or has infinite length. Thus the analogue of Theorem 1 for bounded

functions is false. In the second part (Sections 6 to 10), we prove that for any individual level

* Research of the second author is partially supported by the National Science Foundation.

366

Level sets of univalent functions 367

curve %, we have

13`~1-< A2 < 1018. (1.3)

These two parts make up the bulk of the paper. In the third part (Section 11) we complete the proof of Theorem 1. We show

first by an elementary transformations that (1.3) leads very simply to

I%1 -< 2A28(%). (1.4)

Using this and (1.2) we deduce at once that

13̀1 = ~ t%1-<2A2 ~ 8(%)-< 2 A , A > (1.5)

A similar argument to that leading f rom (1.3) to (1.4) then shows that if E is any subset of 3 ̀ we have

IEI <- 4AIA28(E)

and this gives Theorem 1. This part is relatively short. Our arguments in parts I and I I are based on harmonic measure. If D is a

domain and E is a subset of the closure /3 of D, we denote by co(z, E, D) a function which is harmonic in D \ E and has boundary value 1 on E and zero on O D \ E . Thus if E is a subset of OD, co(z, E, D) is the harmonic measure of E with respect to D at z. Clearly co(z, E, D) increases with expanding E for fixed z and D and with expanding D for fixed z and E. If no confusion is likely because D is fixed we sometimes write simply co(z, E).

2. Elementary lemmas on harmonic measure

The only property of the 3`k which we shall use in order to prove (1.2) is that the % are arcs in A with end points on the boundary of A, which satisfy the separation condition involving harmonic measure which is described in Lemma 1. However , in order to use Lemma 1 we need various other properties of harmonic measure.

L E M M A 1. If 3`k are the components of the level set 3' and 3`'k = 3 ' \ 3 ` k , then

co(z, 3`L A)<�89 z e 3`k. (2.1)

We recall that w - - f ( z ) maps A onto the domain ~. Then 3`~ is mapped onto an arc l of L and 3`[ is mapped onto the remainder l' of L fqO. Clearly we may

3 6 8 W. K. HAYMAN AND J.-M. G. WU

assume that L is the real axis, and (2.1) is equivalent to

to(w, l ' ,O)<�89 t o e l (2.2)

in view of the invariance of harmonic measure under conformal mapping. To prove (2.2) we construct the reflection O* of O in L and define U to be that component of O n O * , which contains I. Write

to(w) = to(w, l ' ,O), to*(w) = to(w)+ to(~);

then to*(w) is harmonic in U. Let ~ be a boundary point of U. Then, since 12 is simply-connected, ~ cannot lie in l'. Thus ~ is either a boundary point of 12", in which case to(~) = 0, to (~) < 1, or a boundary point of O*, in which case to(~) = 0, t o (~ )< l . Thus t o* (~ )< l and so t o * ( w ) < l in U. Taking for w a point on l, we deduce (2.2) and hence Lemma 1.

If O is the region - 0 < a r g z<2"rr-O, where 0 < 0 < - r r , and l and l' are the negative and positive real axes, we have to(w)= ( z r - 0 ) / ( 2 I r - 0) on l, so that (2.1) is sharp.

Our next lemma is a special case of the Milloux-Schmidt inequality [3, p. 109] and [6, p. 107].

L E M M A 2. Let ~q be an arc in AR ={z l l z l<R} , which passes through the origin and has one end point on [z[ = R. Then

to(z, ~1, AR)----- 1 _ 4 arctan ([zl/R) '/2. -/r

We also need a variant of Hall 's Lemma [2]. It is pointed out by David Drasin that a theorem which bears some resemblance to Lemma 3 in disks was proved by Maitland [5].

L E M M A 3. Suppose that H = {z [ z = x + iy, y > 0} is the upper half plane. Let E be a relatively closed set in { 0 < y < a / 1 0 0 } and let E * = { x l x + i y ~ E } be the projection of E on the real axis. Then for Im z >- a,

to(z, E, H)>_2a~(z, E*, H).

As in the proof of Hall 's Lemma we may assume that E is the union of a finite or countable set of line segments lk, which are parallel to the real axis and whose projections have at most endpoints in common. This is also the only case of


L e m m a 3 which we use. Further we assume, as we may, that a = 1. If

lk = {t + ibk, ak <-- t <-- a~}

we define for z e H

1 1

where

G(z, () = log

is the Green function in H. When ~ = t+ ib, z = x + iy, we set I z -KI = r, I z - tl = 0 and assume that y -> 1 >- 100b. Then

1 ~ - , (1 4by \ 1 (1 +~_2y ) - +7)_ log

>__2y//1 4by \ 2 y / [ 1 \ 2 5 2 y + 7 ) > - 7 / k 1 + ~ - ' ~ ) = 26 #2 "

Thus

> 2 5 1 ~ " y 25 U ( z ) - - ~ ~ %| ---~dto =~-~ to(z, E*). (2.3)

Next, we prove, following Hall, that

�9 - 2 U(z) <---~+-~, z ~ n . (2.4)

We recall that

if-z-/_ = G log

The right hand side decreases with increasing b when z and r are fixed and so

3 7 0 w, K, HAVMAN AND J.-M. G. WU

assumes its maximum value for b = y - r if r < y and for b = 0 when r - y . Thus

~ l og I - r

where

M(r ,z )= 1 l o g ( ~ - l ) , r < y y - r

2y M(r, z) = ~-, r->y.

Also M(r, z) decreases with increasing r and so

~-log ~ <--M(r, z) if Iz-r

In particular this inequality holds if ~ = t + ib, where t - x = :~ r. Thus

U(z)<- 1 M(r, z) dr 71"

,og w y - r

The second integral is 2/~. To evaluate the first integral set y - r = ty, then

log T ~ / T

o ( 2 n + 1 ) 2 4"

This proves (2.4). Now the maximum principle yields

U(z)<_(4+2)w(z ,E) , z ~ H .

Combining (2.3) and (2.5), we conclude for y-> 1,

2 - i

and this proves Lemma 3.

(2.5)


L E M M A 4. Suppose that E is a subset of the real axis, that a > 0 and that I is the interval [ - 1 0 a , 10a]. Then if

to(ia, E, H) < 7 / 8

we have

I I \ E I >- ~-~ I/I.

Wri te E ' = I \ E , IE'I = 2n . We suppose wi thout loss of generali ty that a = 1. Then if a ( t ) denotes the measure of E ' N [ - t , t] we have

to(i, E ' ) =--1 I~ 1~ dA(t) l + t 2"

Our hypotheses imply that a ( t )<--2min (t, rl). Thus

to ( i ,E , )<2 ~ "~ dt 2 - - ~ l + t 2= tan-lr l .

Also

7 > to(i, E ) >- to(i, I) - to(i, E ' ) > 2 (tan_~ 1 0 _ tan_trl)" 7/"

Thus

--2 2 _ 2 1 tan- l r l ~ - - t a n - l l 0 - 7 = ~ - --- tan -1 ~ > ~

7I" "tr "rr 5,'tr'

'rr 1 n > i - ~ - ~ > 0.09 > ~ �9

This proves L e m m a 4.

L E M M A 5. Suppose that ~ is an arc in {z I l x l - 32, 0 < y < 1} which connects the lines x = ~:32. Then if l < y o < - 2 , we have

to(iyo, rl, H ) >~.18

3 7 2 w . K . HAYMAN AND J.-M. G~ WU

Let l = 32. Le t rh , "02 deno te the segments { - l + iy, 0 - - < y ----- 1}, {l + iy, 0 - < y -- 1} respect ively. Le t "q3, ~14 deno te the segments { - l - 1 --< x -- - l + 1, y = 0} and { l - l <-x<-l+ l, y=O}. T h e n

oJ(z, ~3)>'�89 on rh and oJ(z, T[4)~ 1 o n ~2-

Thus for z ~ H

~o(z, 7h) + w ( z , 7h)--< 2(~o(z, ~13) + o~(z, '1~4)).

Final ly if r15 deno t e the s egmen t { - l - < x - l, y = 0} then the m a x i m u m principle shows tha t for z = Xo + iyo, where Yo > 1,

~(z, n~)<-~(z, n)+,o(z, nl)+~o(z, n2),

so that

oJ(z, •) -> ~o(z, "qs) - 2(oJ(z, 713) + oJ(z, ~h))-

Set t ing z = iyo where 1 - Yo -< 2, we obta in

o j ( z , ~ l ) > 2 ( l 1+1 _ l 1) - - tan -~ - ~ + 2 t a n -~ Z 2 "rr ~ - 2 tan -1 - -

= 1 ---2 ( tan-1 2 /z---~) 1r 1 + 2 t a n - 1

> 1 ---4 + > ~ .

This p roves L e m m a 5.

3. Construction of segments

T h e inequal i ty (2.1) is the only p rope r ty of the level curves which we use in o rde r to comple te our results. I t is conven ien t to work in H ra ther than A since the noneuc l idean met r ic is easier to visualize in H. W e consider in the first instance only those level curves which lie ent i re ly in the square

R o : 0 - - < x - 1 , 0 - -<y-< l . (3.1)


More precisely we assume that w = F(z) is a univalent map from H into the complex plane, whose image does not contain the whole real axis L in the to plane, and denote by l i those components of F-I(L) which lie entirely in R0. (The simpler case L c F(H) will be considered at the end of Section 11.) If

l~= U l~,

it follows from Lemma 1, the invariance of harmonic measure and the fact that to(z, E, H) increases with expanding E that

to(z,l'k,H)<-�89 on Ik. (3.2)

Apart from (3.2) we only use the fact that the lk are arcs which lie in R0 and have both end points on the segment 0-----x--< 1, y = 0.

It proves convenient to replace each Ik by certain horizontal line segments qk of diameter comparable to the diameter of lk. We can do this at the cost of replacing �89 by 7 in (3.2). Using Lemma 3 we shall deduce that the projection of a suitable subset of Ui~k qi has harmonic measure less than 7 on qk and so by Lemma 4 must leave uncovered ~6o of a neighborhood of the projection of qk. From this we can deduce that the sum of the diameters of the qk and so that of the Ik is bounded by a (very large) absolute constant and (1.2) will follow.

Let m, n be integers such that

n_>10, 0<m___2 ".

We shall call a dyadic segment the set

q ( m , n ) : ( m - 1 ) 2 - " < x < m 2 - " , y = 6 0 0 . 2 - " . (3.3)

The projection

J ( m , n ) : ( m - 1 ) 2 - " < x < m 2 -n, y = 0 (3.4)

of q(m, n) on the real axis will be called a dyadic interval. We note that two dyadic segments are disjoint, unless they are identical. Two dyadic intervals are disjoint unless one is contained in the other. Also all dyadic segments lie in Ro.

L E M M A 6. If lk is a level curve in Ro there exists a dyadic segment qk such that

6(qk) > 10-5~(/k) (3.5)

374 W. K, H A Y M A N A N D J.-M. G. W U

and

~o(z, lk, H ) > ~18 for all z on qk. (3.6)

Let x~, x2 be the lower and upper bounds of x on lk and let h be the upper bound of y on lk. We write x2--xl = 2d and distinguish two cases.

(i) Suppose first that d > 40h. In this case we define n to be the largest integer such that

d 2-" > ~

24 ,000 '

and then define m to be the largest integer such that

m2-" --- �89 +x2).

Let q(m, n) be the corresponding segment given by (3.3), and suppose that = ~ + i~1 ~ q(m, n). Then

d d

Also if Xo=�89 we have for ~=(+ i~1 on q(m, n)

d Ir Xol-< 2-" < (12,000) -1 a <

Thus lk has a subarc l', lying in the rectangle

d Ix-~l<3:a 0 s ! - - 4 o u, < Y <~'~,

and joining the sides x=~w32d /40) of this rectangle, since by hypothesis h---d/(40). Now it follows from Lemma 5, d/40<-q--< d/20 and conformal invariance that

6e(~,/k):=- w(~ , / ' )> i1~9,

which gives (3.6). Further, using our choice of n we see

$(lk) ----- (h 2 + (2d)2) 1/2 <-- 2d(1 + (~o)2) 1/2 < 50 ,000 .2 - " ,


which yields (3.5). Thus L e m m a 6 is p roved in this case. (ii) Suppose next that d <-40h. We choose for n the least integer such that

6 0 0 . 2 - " --< h,

and set rl = 600 �9 2-", so that

h/2 < rl <- h.

Since Ik has endpoints on the positive axis we see that Ik contains a point

~o = ~o + irl. If ~o = 0 we choose m = 1 and otherwise we choose for m the smallest integer such that m �9 2 - " -> r Then if ~ is any point on q(m, n), defined by (3.3), we have

1 ~ - ~ o 1 ~ 2 - " = ~ / 6 0 0 < ~ tan ~ .

Now lk contains a subarc l' lying in the disk D o : l z - ~ol < "~ and joining the point

~o to the c i rcumference of Do. Hence L e m m a 2 shows that for ~ e q(m, n),

~(~, lk, H)>>-oa(~, l', H)>-~((,, l', Do)

4 11'-~ol 4 7r 18 - - > l - - - a r c t a n ~/ .q ->1 . . . . . .

rr rr 76 19"

Also 8(/k) <-- ((2d) 2 + h2) 1/2-< h(1 + (80)2) 1/2

< 8 1 h < 8 1 �9 1200" 2-".

Thus (3.5) and (3.6) are satisfied and L e m m a 6 is proved.

W e now prove the required separat ion p roper ty for the segments qk, which is

the ana logue of L e m m a 1.

L E M M A 7. Let qk be the segments defined in Lemma 6 and write

q~= U qi.

Then we have

o(z, q L H ) < ~ , zeqk.

3 7 6 w . K . HAYMAN AND J.-M. G. WU

We note that

tO(z, lk)+tO(z,l~)<_3, z eH . (3.7)

For if z lies on lk, the inequality holds in view of (3.2) and if z lies on l~, where j ~ k, we have

tO(z, lk) -- tO(z, 1,x < 1

SO that (3.7) still holds. Also the left-hand side of (3.3) is zero on the boundary of H and is harmonic in H except on the I i. Thus (3.7) follows from the maximum principle.

Using (3.7) and (3.6), we deduce that

tO(Z , ir'~..=-_3 18 _ 2 1 ~kl=2 19--38 on qk- (3.8)

Further we have on qi, where j ~ k, from (3.6)

1 = to(z, q~)< ~to(z, lj)<-~o(z, l~).

Thus

s r 19 qk)<~to(z, ID for zeq'k (3.9)

and (3.9) trivially holds on l~ and on the boundary of H. Thus (3.9) holds in H. Combining (3.8) and (3.9), we deduce on qk

21__ to(z, q;3 < ~ " ~-~-~

and this proves Lemma 7.

Let S be a collection of distinct dyadic intervals I on the real axis. Suppose

that al , a2, a3 are positive numbers such that a l - 1, 0 < a2--< 1, 0 < a 3 < 1. If I is an interval I x - Xo[ <�89 III, we write aI for the interval I x - x0l <�89 III. We shall say that S satisfies the hypothesis P(al, a2, a3) if for every I ~ S there exists a set e(I) such that

e(I) c alI (3.10)

l e (~ l - a,a2 I/I (3.11)

Level sets of univalent functions

and if I ' is any interval of S such that II'1 <-a3 II[ we have

e( I) n r = 4,.

377

(3.12)

We shall prove in Lemma 9 that such a collection S necessarily has finite total length. Thus in order to prove that Z ~(lk) is finite, it is sufficient, in view of Lemma 6, to prove that the projections Jk of the qk onto the real axis satisfy P(al, a2, a3) for suitable constants al , a2, a3. We proceed to deduce this result from Lemma 7.

L E M M A 8. The projections Jk of the dyadic intervals qk are all distinct and satisfy P(12,000, ~,1 ~).1

It follows at once from L e m m a 7 that qk N q~, = 4~ so that the different qk are disjoint. Thus their projections are not identical. Suppose that q = qk, and let Zo=Xo+600 �9 2-"i be the midpoint of q. Let J be the projection of q on the real axis and let E be the union of all those qj whose length is less than 10 - 2 . 2 -n.

Then it follows from L e m m a that

~(Zo, E)-< ~(Zo, q~) <~.

Also by our construction E lies in 0 > y > 6 �9 2-". Thus if E* is the projection of E on the real axis, we deduce from Lemma 3 and conformal invariance that

o~(Zo, E*)---~o(z0, E)<~.

I t now follows from L e m m a 4 that if I is the interval 12,000J, i.e. Ix - Xol < 6000- 2 -n, then

I I -E*t---~ IXl = 12,000. ~ IJI.

Letting e(J)= I \ E * , we conclude Lemma 8.

We can now obtain a bound for the sum of the diameters of the lk in Ro by

proving

L E M M A 9. If the collection S of dyadic intervals I satisfies P(al, a2, a3) then

~,lll<_K(al, a2, a3)<_288 a---! (5+log a~21a% ). (3.13) s 32

378 w . K . HAYMAN AND J.-M. G. WU

Thus if lk are the level curves in Ro, we have

y. 3(lk) < - 105K(12,000, 112o, ~_L~loo~ ~'1 ~,,la15. (3.14)

If I is a dyadic interval, then I = J ( m , n) for some m and n as in (3.4); we call n = n(I) the index of L Thus III decreases with increasing n(I).

4. Proof of Lemma 9

We first assume that if J(m, n) and J(m' , n') are intervals of S then

m = m'(mod kl) and n = n'(mod k2), (4.1)

where k~, k 2 a r e the least integers satisfying respectively

kl -> 3al (4.2)

and

1 48a l 2 k2 >-------" -~ (4.3) a 3 a 2

If S does not satisfy (4.1) we shall divide S into at most kxk2 subclasses each of which satisfies (4.1) and sum over each subclass separately.

Let F~ be the subset of all points of (0, 1) which are covered by at least v distinct intervals of S. Evidently F~ is the union of a finite or countable set of intervals in S. Also Fv+l c Fv. Let Io be a component of F~. We proceed to show that

lIonF + l-< 1Iol. (4.4)

To see this, suppose that n(Io) = no and let nl be the least index greater than no of intervals of "S which meet �89 We call V1 the class of all such intervals of index nl. We note that by the first relation in (4.1) the intervals 3a111, where I1 e V1, are disjoint. We now define

\ H1 = � 8 9 3a111.

\ v ~


More generally if the class Vk of intervals Ik with index nk has been constructed we write

Hk=�89 3alI, U U 3a,/2U ,. 'uU 3a,Ik) I Vz Vk

(4.5)

and define Vk+~ to be the union of all intervals of S, meeting Hk and having least index nk+l > no. By our construction nk.l > nk.

This process continues indefinitely or stops, either because Hk is empty or because Hk meets no intervals of S with index greater than no. We then define

H = N H k , V = U V k ,

where intersection and union are taken over all k for which Hk is defined. We now distinguish two cases.

(i) Suppose that IHt ~ �88 lIol. We note that no point of H meets any interval I of S having index greater than no. For suppose contrary to this that I is such an interval of least index n. Let k be the largest integer for which nk < n and Hk is defined. Then I meets Hk and so belongs to the class Vk+~, and thus I is disjoint from Hk+l, contrary to hypothesis. Thus in case (i) H does not meet Fv+l and

tl,,nFv+l[<-IIo\Hl<-]llol.

Since Fv+2 r Fv+I, we deduce (4.4) in this case. (ii) Suppose next that IHl<k I/ol.

In this case we see that

Thus

III >__~1 iio[" (4.6) v 12al

Suppose that I is an interval in Vk. Then by hypothesis alI contains a set e(I) not meeting any interval of S having length less than a3 tII, i.e., no interval of S having index greater than nk in view of (4.1), (4.3) and the fact that I has index nk. We note that the sets e(I) corresponding to distinct intervals I are disjoint,

380 w . K . HAYMAN AND J.-M. G. WU

since the intervals al ! are disjoint by (4.2), (4.5) and the fact a ~ - 1. Thus using (3.11) and (4.6) we deduce

I~ e(I)I = ~v [e(I)l >- a,a= ~v [II >- ~22 Ilol. (4.7)

We have just seen that e(I) corresponding to I~ Vk meets no interval of S having index greater than nk. Next suppose that e(I) meets an interval I ' e S having index n, no< n < nk. It follows from (4.1) that n- - -nk- k2, so that

II'1- 2 k= III.

On the other hand, I ' cannot contain I since otherwise I' would meet Hk-i and have index less than nk, contrary to hypothesis. Thus e ( / ) N I ' must lie in one of two subintervals of I', which adjoin the end points of I ' and have total length

(a~ + 1) II1~ 2a~. 2 -k= IVl.

Thus we see that the total length of the intersections of the e(/) with intervals I ' of S having index n, such that no< n(I')<n(I) is at most

~7~a121-k~ II ' l- a12 l-k2 I/ol, (4.8)

where ~ is taken over all the maximal intervals I ' in S of index greater than no and containing some point x ~ Io. For these intervals I ' are disjoint and lie in Io and so have total length at most Ilol. Thus if e'(I) is the subset of e(I) meeting no intervals of S of index greater than no and different from nk, we deduce, using (4.7) and (4.8) that

l!..J e'(I)[-> tU e(I)l-al" 21-k~ Ilol

-> ( ~ 2 - a 1 2 1 - ' ) I/ol. (4.9)

Using (4.3) and (4.9) we deduce that

I U e ' ( ~ l ~ l I o l .

Also a point x in e'(I) lies in no interval of S having index n > no, n# n(I). Thus xr F~+2. Hence in this case (4.4) holds, so that (4.4) is true in all cases.


We write

a2 0 = 1 - - 24 '

and deduce by induction that since IF~I ~ 1, we have

IF2vl---IF2~_,t--- 0 v - ' , ~>-1.

Again

t 2 ~ l t l = F~I = ( I G v l + l F ~ _ l l ) - < 2 0 v - l = - s 1 1 1 1 - O

Thus with the hypotheses of Lemma 9, together with (4.1) we have

2 48 Y. II[< . . . . (4.10) s - 1 - 0 a2"

Also in the general case, when (4.1) is not satisfied we can divide the intervals of S into at most klk2 subclasses each of which satisfies (4.10) so that we always have

~, III <-48klk2/a2 = K(al , a2, a3). S

Using (4.2) and (4.3) we obtain

kl -< 3al + 1 ---4al,

2 log 2 \ a - ~ 3 / 1 2 ',a2a3/ /

We deduce that

<3 17.1( ( al ) ) ( al ) K - ~ . 4 - 4 8 - - 5+ log = 2 8 8 a l 5 + l o g ~ . a2 ~3 ~2

This proves (3.13). Next it follows from Lemma 8, that if the class S consists of the projections of

the qk, we may take

al = 12,000, a2 = 1-~o, a3 = ~ .

382 w . K . H A Y M A N A N D J.-M. G. W U

Thus

I qk I -- 288- 12,000- 120(5 + log (1.44 x 108))

< 101~

Using (3.5) we obtain

~(lk) < 10 t5.

This completes the proof of Lemma 9.

5. Proo[ o! (1.2)

Let Tk now be the level curves of a univalent function in A. We divide Tk into 3 subclasses. Consider first those Tk, which lie entirely in

al = a n ( x > -fro). (5.1)

We consider the transformation

(1-z !_ i((1-tzl2)-2iy 1 w=.+iv l+zJ+2- iGTzl )+i

Clearly A corresponds to the upper half-plane H in the w plane and the subset (5.1) maps into the unit square Ro given by (3.1) with (u, v) instead of (x, y). The level curves Tk in At correspond to level curves Ik in Ro. Also

=511+z12>~~ in ,t 1 ,

and so

in the image of A1. We deduce that

8(Tk) < i08(/k).


Using Lcmma 9, we obtain

E l~('yk) ~ 1016. (5.2)

Similarly if Az = A n ( x < ~ ) and Y.2 denotes the sum over those 3'k which lie entirely in A 2 we obtain

~".2~ ('yk) "< 1016. (5.3)

Consider now the remaining level curves 3% Each of them must contain an arc Tk with end points on x = :~ ~ in A and hence meets the imaginary axis at finitely many points iyk with [ykl< 1. We choose the least such Yk, and enumerate the Vk in order of increasing Yk. We proceed to prove that if Yk-----0 then

Yk+l -- Yk > 1 . (5.4)

Suppose that (5.4) is false for some k. We note that 3'k separates iyk+l from the arc.

~ o < x < ~ , y = - ~ / 1 - x 2 (5.5)

in - 1 < x < 1 . Let Ak be the disk

]z -- iyk ] < ~o

and let A~, be that component of Ak \~/k which contains "iyk+l. Then A~, cannot contain any point of the arc (5.5). Thus since Yk--<0, we deduce that A~,__qA. Hence if ~k = Tk r Ak, we deduce that

co (iyk+l, g/K, A ~) -< co(iyk+I, g/k, A) --< co(iYk+x, Tk, A) --<�89

in view of Lemma 1. On the other hand g/k contains an arc "0k joining z = iyk to the boundary of Ak and so we deduce that

co(iYk+x, "ok, Ak) <- co(iyk+l, "~k, Ak) = co(iyk+I, g/k, A~) --<�89

Now we apply Lemma 2 and deduce that

1 yk+l-- yk-->~-o tan z ~ > ~ .

384 w.K. HAYMAN AND J.-M. G, WU

Thus (5.4) is true after all and less than 70 different ~/k can meet the ncgative

imaginary axis. Thus there are at most 140-/k in our remaining group and if ~3

denotes the sum over these, we have

Y~3 8(~/k)< 300.

On combining this with (5.2) and (5.3) we deduce (1.2).

6. Pre|iminaA'y reductions

We now embark on the proof of (1.3). We confine ourselves to the following special case to which the general result can easily be reduced. We assume that O is the interior of an analytic Jordan curve F and that I is a segment [bl, b2] of the real axis in I2 whose endpoints bl, b2 lie on F. We denote by ~/ the image of I under the conformal map

z = F ( w ) = f - ~ ( w )

of O onto A and shall show that ~/= ~/k satisfies (1.3). In this part we work with the geometry of O rather than that in H or A. If w is a point of O we write

a(w) = inf Iw-~l (6.1) s

for the distance from w to F. We start by constructing a function d~(u) which is comparable to d(u) on I but behaves in a smooth manner. We shall then dissect the interval I into a sequence of intervals/j,k, and with each// ,k we associate an arc Fj,k of F, such that length of the image of//,k by F(w) is comparable with that of Fj, k. The Fj. k will be disjoint and so their images have total length at most 2zr

and f rom this (1.3) will follow. We set w = u + iv. By our construction the line u = c meets F for bl <-c-< b2 and we write

cbo(u)=inf {Iv[lu+iv~F}, bl <-u<-b2. (6.2)

Further we define

~b(u) = i ~ {~bo(Ul) +~z lul - u[}. (6.3)


Clearly 4~(u) ->0, with equality only at the endpoints bl , b E of I. A point u ~ I will be called a spike-point if

~(u) = ~o(U).

Evidently the endpoints bl, b2 are spike points. In order to establish our results we proceed to subdivide I into intervals

bounded by spike points. We prove first

L E M M A 10. I f u is a point o f I and d(u), &(u) are defined as above then

d(u)>-~6(u).

Suppose that Wl = ul + iv~ is a point of F such that

l w , - ul = d(u) .

Then

6(u)-< 6o(U,)+~ lu,- ul-Ivlt+~ lu,- ul ~- [1 + (~)~),j2 Iw , - ul ~- ~d(u).

This proves L e m m a 10.

L E M M A 11. W e have for Ul, u 2 e I

I ~ ( u , ) - ~(u2)l <- ~ lu - u21.

Suppose that u3 e I. Then

~o(U3) +~ lug- u31- ~o(u~) +~ tu2- u~l +~ lu,- u~l.

Taking lower bounds for varying u3 we obtain

~(u , ) -~ (ug+~ lul- u~l.

Interchanging ul and u2 we obtain

6(u9-< ~(u,) +~ lu,- u21

and so we deduce L e m m a 11.

386 w , K. H A Y M A N A N D J,-M. G. W U

L e m m a s 10 and 11 are not t rue if we use ~bo in p lace of ~b.

L E M M A 12. The set of spike points is a closed non-empty subset of L

Since ba, b 2 a r e spike points , the set is cer ta inly non -empty . Next we note that since F is closed the funct ion ~bo defined by (6.2) is lower semi-cont inuous . Also we see by L e m m a 11 tha t ~b(u) is cont inuous. Thus h(u) = ~bo(U)- 4~(u) is a lower semicont inuous nonnega t ive function. T o see tha t h(u)>-0 we just set u, = u in (6.3). H e n c e the set where h(u)<-0 is d o s e d and this is the set of spike points.

I t follows f rom L e m m a 12, that the c o m p l e m e n t in I of the set of spike points consists of a finite or countab le set of open intervals J. In each of these intervals 4~(u) has a par t icular ly s imple form.

L E M M A 13. Suppose that a, a' are spike points in I such that a < a', and that the interval (a, a') contains no spike points. Then for a < u < a',

d~(u) = min {~bo(a) + ~ ( u - a) , ~bo(a') + ~ 2 ( a ' - u)}.

Since a is a spike po in t we have for a~<a, a ~ I ,

6o(a,) + A ( a - a,)-> 6o(a).

Thus for a < u < a ' , we deduce that

~bo(aO +~z(u - a l ) = ~o(a) +~2(u - a ) + 6o(a , ) - d~o(a) +~2(a - a , )

~bo(a) +~z(u - a). (6.4)

Similarly if a 1 ~> a ' , we deduce that

4~o(a,) +~2(a , - u) -- 4,o(a') + ~ 2 ( a ' - u). (6.5)

Next we no te that in the definit ion (6.3) of ~b(u) we m a y allow ul to range only over spike poin ts in I. For since ~bo(U) is lower semicont inuous the inf imum in (6.3) is a t ta ined for some u~ in I. If ul is not a spike point we can find u2 such tha t

4,o(U9 +A l u : - ud < ~o(U,).

Thus

4,o(U9 + ~ lu - u~l-~ 4,o(U9 + ~ lug- ud + ~ lu - u,I

< ~o(U,) + ~ lu - ud =,~(u)


and this contradicts the definition of tb(u). Hence if a < u < a ' there exists a spike

point u~, such that

4 , (u ) = 4 , o ( U , ) + ~ lu, - ul. (6.6)

In view of (6.4) and (6.5) we may suppose that a <-u~<--a ', so that u~ = a or

Ul = a ' , since (a, a ' ) contains no spike points. Using L e m m a 11, we see that Ul is that one of a, a ' which gives the smaller value of ~b(u) in (6.6).

7. A dissection of the interval I

Suppose first that the interval I contains no spike point o ther than the end

points b~, b2. In this case we write al = b~, a2 = b2 and deduce f rom L e m m a 13 that

, b ( U ) = l ~ m i n ( a - a l , a 2 - a ) , a l < a < a 2 . (7.1)

Thus we deduce f rom L e m m a 10 that in this case

2 min (a - al, a2 - a) <-- d (a) , al < a < a2. (7.2)

Suppose next that I contains at least one spike point ao, such that b~ < a o < b 2 .

Having chosen ao we define o ther spike points a t inductively as follows. If a i has been defined, j-> 0, we define ai+l to be the smallest spike point such that

aj+l --- a t + 6q~ (at). (7.3)

We deduce f rom (6.3)

q~0(b2) + 1 l b2 - at l -> 4' (at)

i.e.

b 2 - a i :~. 124f fa i ) ,

so that at+t < b 2. Thus ei ther at some stage at+l = b2, in which case we stop the

procedure , or else the a t are defined for all positive j and

a j - * b 2 , as j - - * + ~ .

3 8 8 w . K . HAYMAN AND J,-M. G, WU

Similarly if, for some nonposi t ive ], a i has been defined we define aj-1 to be the largest spike point such that

0~_ 1 ~ a j - - 6~b (ai). (7.4)

If ai > bl, we deduce again that ai_ 1 --- bl. T h e re levant proper t ies of ou r subdivision are given in

L E M M A 14. The interval I can be divided into a finite or countable set of subintervals [ai, a~+~], where the aj are spike points with the following properties

aj+l - aj - 4 max {$(aj) , 4,(a~§ (7.5)

Further, if d(a) denotes the distance of a from the boundary F of D, we have for a~ <a <ai+l

d(a) >-~ min {max [6(a~), �89 - ai)], max [6 (a j+0 , �89 - a)]}. (7.6)

If a i = b~, aj§ = b2, (7.5) is trivial and (7.6) follows from (7.2). Thus we may assume that a0 is a spike point in I and that the remaining a i are defined by (7.3) and (7.4). We concent ra te on (7.3) and ] - - 0 for definiteness. T he case j < 0 is similar.

We first p rove (7.5). Suppose first that r247 Then (7.5) follows f rom (7.3). On the o the r hand if

4~(aj)<Z6(aj+0 so that [ 6 ( a j + l ) - 6 ( a j ) l > ] 6 ( a j + l )

we deduce f rom L e m m a 11 that

laj - a~+~l----- 1 2 1 6 ( a j + O - 6 ( a j ) [ > 4 6 ( a j + O .

Thus (7.5) holds in all cases. Next we p rove (7.6). Suppose first that

ai <a <-ai +64,(aj).

Then L e m m a 11 shows that

1 6 ( a ) - ~ , (a~) l - - - - -~(a - a i) ---< � 8 9 (a~-).


Thus in this case

,b(a) >-�89 (ai) >--~(a - ai). (7.7)

Using Lemma 10, we deduce (7.6). Suppose next that a i +6~b(a i) < a < a~§ In this case it follows from (7.3) that

if b is the largest spike point such that b <--a, we have

a~ <-b <ai +6ek(aj).

Also the interval (b, a~+l) contains no spike point and so by Lemma 13, we

have

~b(a) = min {4~(b) + ~ ( a - b), ~b(ai§ + ~(ai§ - a)}. (7.8)

In view of (7.7) applied to b instead of a, we have

4,(b) -> �89 - ~ (b - ai).

Thus

~b (b) + ~ ( a - b) -> �89 (ai) + ~ ( a - b) -> ~2(a - aj).

Hence (7.8) yields in this case

d,(a) -> min {max [�89 ~ ( a - ai)], max [�89162 ~(aj§ - a)]}.

Because of (7.7), this inequality also holds for ai < a < a/§ Using Lemma 10, we deduce (7.6). This completes the proof of Lemma 14.

Having obtained the points a~ satisfying the conditions of Lemma 14, we now proceed to a further subdivision as follows. Let (a~, aj§ be one of the intervals

defined in Lemma 14. We write

aj,o = �89 + a~+l). (7.9)

W e then define ai.k for positive k inductively by

_ 1 + aj.k+l_:(ai+l ai,k) ' (7.10)

3 9 0 w . K . HAYMAN AND J.-M, G. WU

prov ided that ai.k+l, so defined, satisfies

aj, k+l -< a i + l - ~b(ai + 1). (7.11)

Otherwise we set this value of k + 1 equal to k2 and define

aj.k2 = aj+l.

If an+a = b2, so that ~b(Oq+l)=0, the process cont inues indefinitely and we set k2=oo. Thus (7.10) defines ai,k+l for 0 < k + l < k 2 .

Similarly we define ai, k for negat ive k inductively by

a~,k-1 = �89 + aj, k ). ( 7 . 1 2 )

T h e process cont inues as long as ai, k-~ so def ined satisfies

a~.k_ t >" a~ + d~(a i ) .

Otherwise we set k - 1 = ka, and define

aj ,k , = a i. (7.13)

In this way (7 .9)-(7 .12) define ai.k for ka < k < k2. W e deduce f rom (7.5) tha t

-oo_< k 1 - < - 2 (7.14)

and

2_< k2_<+oo

W e define

/j,k = (ai, k, aj, k+l),

If kl is finite we set

//,k = (ai, k-1, aj, k+l),

and if k 2 is finite we set

//,k = (a/.k, aj.k+2),

(7.15)

k = k 2 - 2 . (7.18)

k = k I + 1, (7.17)

kl + 1 < k < k 2 - 2 . (7.16)


Thus Ii, k is defined for kl < k < k 2 - 1 , i.e. certainly for k = - 1 , 0. Also the Ij, k cover I apart from isolated points.

8. An association of arcs ~,k of F w i t h Ii, k

Suppose that k I < k-<0. We construct an arc of the circle

Iw-a~l=la~.k-ajl (8.1)

starting from the point aj.k. If ai is the left end point of I we start anticlockwise into the upper half plane. Otherwise ~bo(ai)> 0 and one of the two points

a i + irko(a i) (8.2)

lies on F. If a~ +iebo(a i) lies on F, we start the arc of (8.1) by moving in the anticlockwise sense into the upper half plane; otherwise we move in the clockwise sense into the lower half plane. In either case we continue along the circle (8.1) until we first meet a point of F which we denote by b~.k. Since F contains points inside or on the circle (8.1) namely one of the points (8.2), and since aj.k lies inside F the point bj.k certainly exists.

If kl < k < 0 , the points bj,k, bi, k+~ determine two arcs of F. We choose that arc Fj.k which we reach first when going along the circles

Iw-ail=r, Ioq.k-ajl<r<laj.~§

from the point a i + r in the anticlockwise or clockwise sense according as a i + irko(a i) does or does not lie on F. Thus we have associated with each interval /j,k defined by (7.15) or (7.16) an arc F~.k of F if kl < k < 0 . It follows from (7.13) that at least one such k exists. The corresponding intervals Ij.k cover the interval

[% aj.o]. We proceed in an exactly analogous manner with the intervals

/i.k, 0 -< k < k 2 - 1 . We go along the circle

I w - a j . l l = laj, k - a j . d (8.3)

where 0_<k < k2 starting at the point oq, k until we meet F at bi',k. We then associate with the interval /j,k, 0 - -<k<k2 - 1 , one of the arcs [b~.k, bi'k+l] of F, which we denote by Fj,k. We move along circles I w - a j + l l = r , lai, k +~ - aj+~[ < r < laj.k - ai+~l in the clockwise sense if ai+ 1 + i4)(aj+l) lies on F and

392 W. K. H A Y M A N A N D J.-M, G, W U

the anticlockwise sense otherwise and F/.k contains the first point of F we meet in this way. In general U F/,k is a proper subset of F.

It follows from the construction that our interval I has been subdivided into a fn i te or countable set of subintervals/~.k which are associated with arcs Fj.k of F, and no arc of F is associated with more than one distinct interval of /. We complete this section by proving that in this association distinct arcs are disjoint except for endpoints. We show in the next section that the length of the image of /~.k is not much greater than that of F~,k. From these two facts (1.3) will follow.

L E M M A 15. The arcs Fi, k defined as above are pairwise disjoint except for endpoints.

Let /3j.k denote the circular arc from aj, k to bi, k, o r b[k defined as above. We show that distinct arcs /3j, k are disjoint except for endpoints.

Let /3 be one of these circular arcs starting at a = ai,k. Then /3 can contain a semicircle s starting at a only if the other endpoint a ' of s lies outside I. For if a ' lies in I, then the segment aa' together with s constitutes a closed Jordan curve c in /2 . If aj is the midpoint of aa', then our construction ensures that one of the points a~i~b(a i) lies on F and so is in the same halfplane as c and so by (7.11) inside c, which is impossible since /2 is simply connected.

Suppose now that /31,/32 are two distinct circular arcs starting at P1, P2 and first meet ing at a point P o f /2 . Consider the curve 3" formed by going along/31 from P1 to P then along/32 f rom P to P2 and returning along the segment P1,/)2- By construction 3, lies in /2 and the arcs P1P, PP2 have only endpoints in

common. Neither/31 not/32 can mee t the segment P1P2 again since/31,/32 have no points in I other than P1, P2 respectively. Thus 3' is a Jordan curve in /2 .

We shall show that 3" contains a point of F in its interior and this leads to a contradiction s ince/2 is simply connected. Suppose that P1 lies to the left of P2- It is not possible for the centres of both circular arcs to lie outside the segment P1P2, for if the centres are on the same side, the circles are concentric and distinct (1~ and can certainly not meet , and if they are on opposite sides, P1P2 is the shortest distance between the circles. Suppose then that at least one centre, say a i, the centre of the arc P1P, lies on PIP2. Suppose also that zj = aj + ieb(aj) lies on F. We

distinguish a number of cases.

(i) Suppose first that neither/31 nor/32 contains a semicircle. Then/31,/32 both lie in the upper half plane. We show that in this case the segment (a t, z~) lies inside 3". Suppose first that P2 lies inside the circle sl, of which/31 is an arc. Then the

1 If P = aj is the centre of the arc at P2 = aj.k and P lies to the left of, P1, then, by (8.1), P1 = aj,k', where k '<k<0 .


centre of the circle having/32 as an arc must lie to the right of aj and so to the right of P2 since otherwise/32 would lie inside sl. Hence the segment [aj, zi] does not meet /32 and so zj lies inside 3'.

Thus P2 must lie outside sl and so does the whole arc /32, since if /32 went inside sl, /32 would contain a semi-circle. Hence in this case the interior of 3' includes all points in the upper half plane and inside sl and so in particular zj because of (7.11).

(ii) Suppose that /31 contains a semi-circle, but 132 does not. Then /32 lies entirely in the lower half-plane and again the interior of 3' contains all points in the upper half-plane and inside Sl, and in particular zi.

(iii) Suppose that/32 contains a semi-circle but that/31 does not. If the centre ak of the circle s2 containing/32 lies to the right of P2, then the whole of s2 lies to the right of P2, and so cannot meet the segment [ai, zi]. Thus in this case z~ again lies inside 3'. If on the other hand ak lies on P1, P2 then we have the case (ii) with P1, P2 interchanged. Finally ak cannot lie to the left of P1, since otherwise/31,/32 would be arcs of concentric circles which cannot meet. For all circular arcs starting from a point between ak and P2 have centre ak.

(iv) If /31, /32 both contain semicircles, they must reduce to semi-circles, since otherwise they would have two distinct points of intersection. In this case /31,/32 meet at P, which is to the right of P2 and again the point z i lies inside 3".

Thus in all cases /31,/32 can have at most end points in common, since otherwise 3' contains a point of F in its interior, which contradicts the fact that 3' lies inside F.

Suppose now that Fhk is an arc corresponding to an interval Ii, k. Let/3k,/3k§ be the arcs of the circles (8.1) if k < 0 , or (8.3) if k->0, to the points aj,k, aj,k§ Then/3k,/3k§ and the interval [aj.k, a~.k§ determine a crosscut 8j,k in O. In view of what we have just proved distinct crosscuts 8j.k may have a common arc/3k or a common point aj,k, but cannot cross each other. Thus if Dj.k is the interior of the Jordan curve formed by 8j,k and Fi. k then two distinct domains Dj, k are disjoint.

In fact otherwise one of these domains would lie inside the other. However our construction ensures that the interval I lies outside all the Di,k, since none of the /3k meet I again. Hence points near/~,k inside Dj.k are exterior to all the other domains D;,k,. Thus Di, k cannot lie in Dr,k,. Now it follows that two distinct arcs F/,k are disjoint except for end points and this proves Lemma 15.

9. Images of intervals and associated arcs

Suppose that/j,k is an interval of the real axis and that Fi,k is the associated arc of F. We assume for definiteness that k < 0. If k -> 0 the argument is similar. We

394 W. K. H A Y M A N A N D .1.-M. O. W U

recall from (7.9) to (7.11) that

Oq. k =aj+2(k-1)(aj+l--aj), kl<k<_O. (9.1)

We denote by to(w) the harmonic measure of Fi.k with reference to the full open set O. We write

r = 2k-1/2(ai+1 -- at) (9.2)

and prove

L E M M A 16. / [ wl = a t +2%, w h e r e - ~ <-6 <- 7 , then t o ( w 0 > e x p {-30.9}.

Let /3 be the arc of the circle (8.1), joining ai. k to an endpoint/3j.k of Fj.k. Le t /3 ' be the corresponding arc of (8.1) with k + 1 instead of k, which joins ai,k+l to the

other endpoint /3i,k+1 of Fhk. Then /3, Q,k, /3' and the segment [at.k, ai,k+l] form a subdomain Di, k of g2. We now define a domain D ' = Di'k as follows. We define 0o

by

sin (�89 = ~ . (9.3)

Then it follows from (7.6) of Lemma 14, that the sectorial region

2-1/Zr < Iw - a i l < 2V2r, larg (w - ai) I < 0o (9.4)

does not meet F and so lies inside ~. Next it follows from our construction that Fhk contains an arc ~lj, k joining the circles I w - ajl = 2:~1/2r in the annulus

2-1/2r < [w - a~l < 21/2r. (9.5)

We now define D ' to be a subdomain of the annulus (9.5) determined by such an arc ~i.k and one of the rays arg ( w - oq)= ~:0o and having the arcs/3,/3' as part of its boundary. In other words if/3,/3' start off in the clockwise sense we choose the ray arg ( w - a i) = +0o and otherwise the ray arg ( w - a i ) = - 0 o . We continue along [ w - a~[ = 2~: 1/2r until we meet the first arc ~kk-

We note that D ' constructed as above lies in the annulus (9.5) and contains the region (9.4). Let to'(w) be the harmonic measure of ~j.k at w w.r.t. D' , and let D1 be that component of D ' N / ] which contains the segment (ai,k, aj,k+l). Then

to(wO>-to'(wO, Wl~DI . (9.6)


TO see this let r be any boundary point of D1. If s lies on one of the circles

Iw - a i l = 2=~l/Zr or on the ray arg (w - aj) = :g0o, we have

,o ' (~) = 0 <-- o~(~)

A n y o ther bounda ry point ~ of DI lies on ~',k so that

1 = ~o(r > os

Thus oJ(w)-~o'(w)>--O on the bounda ry of D1 and so in the interior of D1 by the

max imum principle. This proves (9.6). A further application of the max imum principle now shows that oJ'(wl)

assumes its lower bound for variable ~li,k in the (limiting) case, when D ' reduces to

the subdomain

-Oo<arg(w-aj)<2w-Oo, 2-1/2r<tw-ail<21/2r (9.7)

of the annulus (9.5) and "qj.k to the arc arg(w-ai)=2ar-Oo. It remains to est imate the cor responding harmonic measure. To do this we use the invariance of

ha rmonic measure and set

s = o" + i'r = log (w - aj). (9.8)

We write

<r o = log r,

and note that the cut annulus (9.7) corresponds by (9.8) to the rectangle

AI : O-o- ~ log 2 < o" < O'o +�89 log 2, -Oo<r<2rr-Oo.

W e have to est imate the harmonic measure w.r.t. A 1 of r = 2 r r - 0 o at

sl = log wa = cro + ~ log 2. By the max imum principle this harmonic measure is greater than that of the

two rays

r>_2w-Oo, o- = tr0q:�89 log 2 (9.9)

w.r.t, the half strip

A 2 : - - 0 o < ' r < + % O'o--�89 log 2 < o" < O'o +�89 log 2

396 w.K. HAYMAN AND J.-M. G, WU

at &. This latter harmonic measure is calculated explicitly as follows. We set

z = exp l log 2 l = x + iy.

This maps A 2 onto the semidisk

T2: Iz[< 1, x > 0 ,

the pair of rays (9.9) onto the segment

x = 0, ly]-- < -q = exp (-2~r2/log 2), (9.10)

and sl onto

f-~rOo+, l z I =exp ~ tsar} = Xl+ iyl = qe '~,

say. The harmonic measure of the segment (9.10) at z = x + iy w.r.t. T2 is

1 { tan_~y+r l+ tan_ l " O - y tan_l xrl x'o } 601(Z) = - - tan-1 ,r x x i - n y l+~ly "

For clearly {ol is harmonic and bounded in TI, and O}l(Z)= 1 on the segment (9.10) and {ol(z) = 0, elsewhere on the boundary of T1. To see this when Izl l = 1,

we use the fact that the triangles 0~lZ and 0Z~1-1 are similar in this case. Using the addition formula for the inverse tangent we obtain finally

1 tan- 1 {(r~ 2~x1(1 + ~12)(1 - r~) O91(Z1) 7r -~2)(1 - n2r]) + 4 x ~ 2 J

since ~l < r~ < 1. In our case we have, using (9.3),

{-~r0o~ q = e x P \ l o g 2 } , where 0 .08<0o<0 .081 ,

while ~1 < 10 -1~ Also xl = rl cos 8,r >- rl cos 7~'/16 = r~ sin (Ir/16).

2r~Xl(1-2 2 r l ( 1 ) rr to(z1)>0.999 rrr 2 rD->O'999--~r - r l sin--16

rr 2n ~ > exp {-30.9}. > 1 6 ~r


This completes the proof of Lemma 16.

We must extend Lemma 16 to obtain a bound for to(w) on the whole of//,k.

This is

L E M M A 17. We have in Ii, k

to(w1) -> As : e-37.

We write

w~ =ai+t .

It follows from (7.6) that if

rb(a~)<t~(ai+l-ai) (9.11)

then

2t d(wO>-~.

Also the function co(w) is positive and harmonic in the disk } w - wt] < d(wO. Thus

Harnack ' s inequality [3, p. 64] yields

Idw(wOl< d@wO to(wO

i.e.

(9.12)

hence if wl = aj + q, w] = aj +t~ are two points in the range (9.11) we have

]log w(wa) - l o g to(w~)l-----25 Ilog t~ - l o g tll.

We suppose that

t 1 = 27/1%'< t~ ~ 21/2r or 2-1/2r----- t] < ta = 27/16r

398 w . K, H A Y M A N A N D J.-M, G, W U

and apply Lemma 16 to wl. Thus we obtain

log co(w~) _> -{30 .9 +~6 log 2} > - 3 2 , ai. k - w~ -< a~.k+a. (9.13)

This yields Lemma 17 if ka + 1 < k < 0 . If k = kl + 1, we have to estimate ~0(Wl) also on the interval I ' = [a~, ai.k]. In

this case it follows from our construction that

ai, k ~ a j + 2~b(a~).

Thus if wl is any point on / ' it follows from Lemmas 10 and 11 that

d(a) >-~4~(ai).

We deduce from (9.12) in this case that

? ) , on , ' aw~ I ~ta~)

Thus if w~ is any point on I ' and w~ = aj, k is the right endpoint o f / ' , we deduce that

2 5 log o~ (wl) -> log ~o(w~) - 2r (ai) ~ _ log r 1) - 5.

On combining this with (9.13) we deduce that the inequality of Lemma 17 holds on I ' also and so on all of/i,k. Thus Lemma 17 is proved whenever k < 0. The case k-> 0 is similar and our proof is complete.

10. Proof of (1.3)

We need a final estimate.

L E M M A 18. I f l, )t are the lengths of the images of an interval I' = Ihk and the associated arc F' = I'i,k respectively in the z plane then

l 1 -< - (5 +25 log 2). )~ "/rA 3


Suppose that Wo is any point on I ' and let Zo = pe i~ be the image of Wo in A. Let 3' = [~bl, ~bl + h i be the image of F' on Iz[ = 1. Then, since harmonic measure is invariant under conformal mapping,

1 [*'+~' (1-p 2) d~b ~~176176176176 1 _ 2 0 cos (O_r

h l + p 27r l - p "

Thus

h h l - p _ <

"trr (W 0, /" , aQ) "trA 3

in view of I~mma 17. On the other hand if z=F(w) maps I2 onto zl, then F(w)

Iw-Wot <d(wo) into za and now we deduce from Schwarz's Lemma that maps

1 - 2h iF,(wo)l_< IZol2<2(1-o)< d(wo) -- d(w;i --TrA 3 d(wo) "

Thus

l=I , IF'(w~176 ~rA32A [, d(wo)ldw~ (10.1)

Now we again use (7.6). If I ' is the interval [aj.k, aj.k+l] where k l + 1 < k < 0 , we set

Wo = ai, k + t,

and deduce from Lemma 14 that

a(Wo) >-- at.

Thus

I I d w ~ d t=251~ , d ( w o ) - 2 , t 2

4 0 0 W, K, HAYMAN AND J.-M. G. WU

If k = k ~ + l we must add to [aj, k, aj.k+~] the interval [ai, aj, k]. In this case t -< 2~b(ai), d(wo) -> 4~b(afl and

d(wo)-4~b(a~ ) dt <- ~"

Thus in all cases

~, ldw~ < ~+~log2. d(wo) --

Hence (10.1) yields

l 1 - < (5 + 25 log 2). A --TrA 3

This proves Lemma 18.

Now (1.3) follows at once. For the length L of the level curve is the sum of the lengths li.k of the images of the/i,k. This yields

L < Z l,, k 1 --< 7rA----~ (5 + 25 log2) ~ hj,g < -- ~ (10 + 501og 2) < 1 0 TM

which is (1.3).

11 . P r o o f o f T h e o r e m 1

We can now put our various results together. We need a final Lemma.

LEMMA 19. Suppose that E is a set in A. Then there exists a bilinear function

z = L ( Z ) mapping IZ[ < 1 into A and a set E ' in IZl < 1 onto E, such that

I L ' ( Z ) I < 2 8 ( E ) on E ' . (11.1)

Suppose that the upper hound of Iz[ in the closure /~ of E is O. We suppose without toss of generality in this case t ha t /~ contains z = p so t ha t /~ lies in

10 -z l< -& (11.2)


where 8 = ~(E). We define z = L(Z) by

Z + tr z - or p 2p z = P t+rZ , Z = t p _ r z , where r=p+1$ , 2 p + 8 < t - < l .

If O = 1, we choose t = 1, so that z = L ( Z ) is a bilinear map of [ZI < 1 onto I z l< 1, and the inverse image E ' of E lies in I Z I < 1. If 0 < 1, we choose t just less than 1. Then IZ I - - t corresponds to I z l - O and so if t is sufficiently near 1, IZ I < 1 corresponds to a subset of I z l< 1, which contains Izt<-O and so E.

Then if z e E, so that (11.2) holds, we compute the derivative of L -~ and find

IL , (Z) I = 1 0 - r z l 2 I o - r o + r ( o - z ) [ 2 (O(1-r)+ar) 2 480 <28 , t O ( 1 - r 2 ) - t O ( 1 - r 2) --~ tO(1 - r 2) = r + ~)

which proves (11.1). This proves Lemma 19.

We recall that we have proved (1.3). We now apply this result to

g(Z) = /{L(Z)} (11.3)

where E is a single level curve of f (z) and E ' = L-I(E) . Then E ' is part of a level curve of g(Z) and so

IE'I ~ m2.

In view of (11.1) we deduce that

IEI -< 28 IE'I -< 28A2.

where 8 = 8(E) and this is (1.4). Using (1.2) we deduce (1.5), for the length of any level set ~.

Finally suppose that E is part of a level set of f (z ) and that 8(E) = 8. We again employ the subsidiary function g(Z), given by (11.3) and define E ' = L - I ( E ) . Since E ' is part of a level set 3~ of g(z), we can now apply (1.5) and deduce that

IE'I <- 13'1 <- 2A1A2 �9

Since also E = L(E') we deduce from (11.1) that

IEI-- IE'l <-4A1A28(E).

402 W. K, H A Y M A N A N D J,-M. G. W U

This proves (1.1) with

Ao = 4A1A2 < 1035,

as stated. We have assumed throughout that O is an analytic Jordan domain and that a

level set E is the inverse image of the real axis by F(z) . If E is the inverse image by F(z ) of a circle or straight line L, we can find a bilinear map W = q~(w) which maps L onto the real axis so that E is also the inverse image of the real axis by

f(z) = ~ { F ( z ) } .

Next if F(z) is a general univalent function and E is the inverse image of the real axis, suppose first that the image of E does not cover the whole real axis but leaves out a point Wo. Then ( F ( z ) - Wo) - t is a regular univalent function with the same level set E. Thus we may assume that F(z) is regular and univalent in this case. We now apply (1 .2 )and (1.3) to the level sets E o of F(pz), where 0 < 19 < 1. Clearly F(19z) maps A onto an analytic Jordan domain Also I/3ol tends to IEI as 19--> 1, so that (1.2) and (1.3) also hold for F(z) .

Finally if the image of E covers the whole real axis, then E must consist of a single closed Jordan curve in A. Let P be the upper bound of z on E. Then E has at least one point on Izl = 19.

Let zl = oe *~ be such a point, and write t =�89 +19), Zo = (19- t)e ~~ and

f ( z ) = F(zo+ tz).

Then the level set E ' of f consists of a single curve going from e i~ to e i~ in [zl < 1

and having length IEI/t > IEI. Thus we may apply (1.3) to E ' and obtain

IEI<IE'I~A=

in this case also. Thus (1.3) holds in all cases. Also (1.2) is trivial in this case, since E is connected. Thus (1.2) and (1.3) hold in all cases and so do (1.4), (1.5) and (1.1). This completes the proof of Theorem 1.

We are most grateful to the referee, David Drasin, for his many helpful suggestions on exposition.

REFERENCES

[1] GEHRING, F. W. and JONES, P. W., Oral communication. [2] HALL, T., Sur la mesure harmonique de certains ensembles. Arkiv. Mat. Astr. Fys. 25A (1937), No.

28, 1-8.


[3] HE1NS, M., Selected topics in the classical theory of functions of a complex variable. Holt, Rinehart and Winston (1962).

[4] JONES, P. W., Bounded holomorphic functions with all level sets of infinite length. Mich. Math. J., 27 (1980), 75-80.

[5] MArrLAND, B. J., A note on functions regular and bounded in the unit circle and small at a set of points near the circumference of the circle. Proc. Cam. Phil. Soc. 35 (1939), 382-388.

[6] NEVANLINNA, R., Analytic Functions, Springer-Verlag (1970). [7] PmANIAN G. and WEn~SMAN, A., Level sets of infinite length. Comment. Math. Helvetici, 53

(1978), 161-164. [8] Wu, J.-M., Harmonic measure and level sets of conformal mapping. J. London Math. Soc., (2), 22

(1980), 263-268.

Imperial College 180 Queens Gate. London SW7 2BZ

University of Illinois, Urbana, Illinois 61801.

Received March 19, 1981

level sets of univalent functions

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