Chapter XII Omitted Values and Normal Families

411

Chapter XI1 Omitted Values and Normal Families The center-piece of this chapter is the following remarkable result due to Friedrich Schottky: iff is holomorphic in D(0, 1) and assumes neither of the values 0 or 1, then f is bounded in D(0, r ) by a bound depending on r and If(0)l only, for each r < 1. [In fact we prove a generalization in which f does not assume 0 and f(”) (for some n 2 0) does not assume 1.1 The principal application is the almost immediate fact that a family of holomorphic functions on a common domain, none of which has 0 or 1 in its range, if bounded at a point, is uniformly bounded in a neighborhood of that point. The compactness theorems of Chapter VII then come into play with astounding consequences. Of course “0” and “ 1 ” here are convenient normalizations: any two distinct complex numbers would serve as well. To scale these peaks we start with some modest preparation of gear.

01

Logarithmic Means and Jensen’s Inequality

Lemma 12.1 The function log+ defined on [0, co) by log+( x )

=

log(max{1, x } )

has the following properties: (i) (ii) (iii) (iv) (v)

log+ is continuous, non-decreasing and non-negative log+(x,. 5 log+(x,) + log+(x,) Vxl, . . ., x, E [O,co) log+(xl . x,) Ilog+@,) . - log+(x,) log(n) vx,, . . ., x, E [O,co) log+(x) I log(1 x) I ZTx for all x 2 o llogxl = log+ ( x ) log+ (l/x)and log x = log+ ( x ) - log+ (I / x ) for all x > 0.

+-

+ + a x , )

+

+ +

+

+

Proof: As the composite of the continuous non-decreasing function x-+ max{ 1, x} of [0, co) onto [ I , co) and the continuous non-decreasing function log: [l, co) -+ [0, a),the function log+ is continuous, non-decreasing and nonnegative. Now (ii) is trivial if its left side is 0. Otherwise, some x j > I and we may suppose the notation chosen so that 0 I X I Ixa 1 * * * x5j - 1 5 1 I x j <.*.I x,.

Then

Omitted Values and Normal Families

412

and so log+(&* *

5 log+(x,*

e x , )

5 log

+

(XI)

*

ax,)

+ + log(x,)

= log(x,. * *xn)= lo&,)

+ - + log * *

+

(x,)

+ + log * *

* * *

+

(x,).

Similarly, (iii) is trivial if each x, < l/n. Otherwise, let the points be monotone in their indices as before and have x, 2 l / n and log+(x,

+ ’ -+ x,) *

*

Ilog+(nx,) = Ilog(n)

+

+

log(nx,) = log(n) log&) log+(x,) * * * log+(&).

+ +

Since log+ is non-decreasing, we have

+ x ) = log(1 + x). Also the function f ( x ) = fi - log(1 + x ) has derivative log+@) Ilog+(l

on (0,m) and so f is non-decreasing. Since f(0) x 2 0, completing the proof of (iv).

=

0, we havef(x) 2 0 for all

For (v) consider the two cases x 2 I or x < I . Definition 12.2 If r 5 0 and f is a continuous complex-valued function on C(0,I ) , define the means M(r,f) = ~qlf(retB)l BE

If, moreover,f is never zero, define

Then by 12.1(v) we see that WYf)

=

WYf)

+ +j).

Theorem 12.3 Iff is holomorphic in D(0, r), 0 c R c r and f has no zeros in D(0, R), then

Proof: Let F be a holomorphic logarithm for f in some neighborhood of b(0, R), apply the Cauchy-Schwarz formula (5.18 appropriately scaled to the disk B(0, R)) to F and take real parts, remembering that Re F = loglfl.

$1. Logarithmic Means and Jensen's Inequality

413

Theorem 12.4 V F is holomorphic in D(0, 1) and 0 < R c 1, then log+IF(Re'e)I Re[zz]dS Refe + z

Vz E D(0, R).

If in addition F is not identically 0, then (ii)

r ~ ( ~ vo , ~ < r) < R. log M(r, F) I R-r +

If in addition F has no zeros on C(0, R), then (iii)

loglF(z)l I -t Iz' L(R, F ) - R - I4 ~

Vz E D(0, R)\F-'(0).

Proof: First consider the case where F has no zeros on C(0,R). Then F is not

identically zero in b(0, R), hence has only finitely many zeros there (5.62). Let these zeros counted according to multiplicity be zl,.. ., z,. We have lz,[ < R by assumption, so each of the functions

-

is holomorphic in a neighborhood of b ( 0 , R). Thus f = F/c$~. .+,, is holomorphic in a neighborhood of b(0, R) and has no zeros in b(0, R). We apply the last theorem to this function, noting that I+,(z)l = I when 1.1 = R:

Since I+,(z)l < 1 for all z E D(0, R), we have logl+,(z)l < 0 whenever z E D(0, R)\F-'(O), so for such z the left side of (1) dominates loglF(z)l and we get the so-called Jensen Inequality :

Recall that log(x)

=

log+(x) - log+(l/x) for all x > 0, to see from (2) that

Now (equation (2) in the proof of 5.20) Refe + z

=

and therefore

R2 - 1zI2 IRele - .Ia

Omitted Values and Normal Families

414

Since log 5 log+, it follows from (4) that

But also log+ 2 0, so the right side of (5) is non-negative and it follows from (2) and ( 5 ) that log+IF(z)l = max{O, loglF(z)l) log+IF(Reie)I Re[ rz]dB Reie + z

Vz E D(0, R)\F-'(O),

giving (i). While (3) and the inequalities (4) lead to (5).

Now consider the general case and a fixed z E D(0, 1). An easy uniform continuity argument shows that the right side of (i), call it O(R), is a continuous function of R E (121, 1). If F is identically 0, then (i) is trivial. Otherwise the zeros of F do not accumulate in D(0, l), hence there are only finitely many in each B(0, 1 - l/n), hence only countably many in D(0, 1). Thus for a dense set of R E [0, l), F has no zeros on C(0, R). Given R E (121, I), pick R, from this dense set with IzI < R, < R and R,+ R. Then log+IF(z)l IO(R,) by the result of the first paragraph and @(R,) --f O(R) by continuity of O. Thus log+1F(z)1 I @(R),which is (i). Similarly L(R, F) is a continuous function of R E [0, 1) and (ii) is already established for the dense set of R for which F has no zeros on C(0, R). Therefore (ii) follows for all R E [0, 1) by continuity. Remark: Notice that (i) also follows from 5.22(i) and the subharmonicity of log+IF1 (5.26(ii)).

Theorem 12.5 Zf f is holomorphic in D(0, l), then L ( r , f ) is a non-decreasing function of r E [0, 1).

Proof: 12.4 or 5.26(ii) shows that log+If1 is a subharmonic function (5.6) and

so the conclusion follows from 5.26(i). Here is an alternative, more prosaic proof. Let 0 < r < R < 1. We may apply 12.4(i) to write

Integrate this :

8 1. Logarithmic Means and Jensen's Inequality

415

Since the integrand on the right is a continuous function of (8, t ) E [ 0 , 2 ~ ]x [0,2~], it is possible to invert the orders of integration in this iterated integral (see 8.26(vi)) to get

But (formula (4) in the proof of 5.20) this inner integral is 1 and so we get

Theorem 12.6 Let F be a holomorphicfunction in D(0, 1) which is not identically

0 but satisfies

Then F has at most countably many zeros. Let them be, with due regard to multiplicity, zl, Za, . . . Then

.

Proof: Since for each R < 1, F has only finitely many zeros in b ( 0 , R), we

may suppose the notation chosen so that lzll I lzal 5 - ... Moreover, by dividing off an appropriate power of z, we may suppose without loss of generality that 0 is not a zero of F. Given n, choose R so that Iz,I < R < 1 and R # lzkl for each k. Let zl,. . ., Z, ( N 2 n) be the zeros of F in b(0, R) and for each 1 Ij I N define

Then f is holomorphic and zero-free in a neighborhood of b(0, R). Apply 12.3 with z = 0, remembering that /+,I s 1 on C(0, R), to see that

Consequently, from (*)

Let R .T 1 through admissible values to have

Omitted Values and Normal Families

416

Since for any x > 1 we have

it follows that

holding for all positive integers n. Remark: The hypothesis (*) is certainly fulfilled if F is bounded, so this result generalizes the final conclusion in 6.9(iii). The equality (**) is known as the Jacobi-Jensen formula. See KOWALEWSKI [19181.

Exercise 12.7 (i) (PRIVAMV [1924a]) Let F be holomorphic in D(0, 1) and satisfy (*)

sup /oa'log+lF(pefe)ldfl = K <

03.

O
Show that F satisjes IF(z)I s e x ~ ( s ( l - ~ zVz~ )E)D(0, 1).

Hints: We may assume that F is not identically 0, the conclusion being trivial otherwise. Then F has only countably many zeros. Given r < 1, there exists then an R E (r, 1) such that F is zero-free on C(0, R). Apply 12.4(ii) to get

Using hypothesis (*), remembering that R > r and focusing only on D(0, r ) , we get

IF(z)l 5 eK/tMr-1 ~ 1 ) )

ZE

Vz E D(0, r)\F-l(O).

The restriction z -$F-'(O) is now unnecessary, so we have IF(z)l 5 eK/(n(r-Izl))Vz E D(0, r ) , Vr < 1. Given z E D(0, I), pick r E (lzl, I ) and let r t I . From the last inequality follows IF(z)l 5 eK/('(l-IZl)). (ii)

Jr

Use (i) to show that the boundedness hypothesis in 7.4 can be weakened to sup,, supr< log+Ifn(reie)ldB< 03 and the conclusion of 7.4still follows.

8 2. Miranda's Theorem

417

Hint: The finiteness of this supremum plus (i) insure that {f n } is locally uni[19241for a related extension of 7.4.) formly bounded in D(0, 1). (See PRIVALOV

62

Miranda's Theorem

This section is somewhat long and a bit arduous. It culminates in a generalization of Schottky's theorem mentioned in the introductory remarks. Here the function f omits 0 but one of its derivatives f ( " ) omits 1. The case n = 0 is Schottky's theorem and that is adequate for all the applications we make here, save one (12.18). Moreover, a self-contained proof of Schottky's theorem is offered in the last section of this chapter. Therefore the reader can skip this section and do 8 7 instead. He can then cite Schottky instead of Miranda in the subsequent sections and get the case n = 0 of all subsequent theorems. As noted, this will secure for him all the applications except 12.18.

Lemma 12.8

f is holomorphic and zero-free in the open set U and F = f ' l f , thenfor every n 2 1 the quotientf(")/fisa sum of n! or fewer terms, each of which is a product of n or fewer factors of the form F ( j )with 0 I j < n.

Proof: By induction on n. Trivial for n = 1. I f true for n = N, consider n = N + 1. We have that f'") is a sum of N! or fewer terms of the form f . F ( j 1 ) .. .F ( b )(0 Ijl,. . .,j , < N,k IN).Since

a sum of N + 1 or fewer terms, each of which is a product of N + I or fewer factors of the form F(') with 0 I j < N + 1, the assertion for n = N + 1 follows.

Theorem 12.9 For each integer n 2 1 there exists a constant A , such that for any f which is holomorphic and zero-free in D(0, 1) L ( r , F ) IA ,

+ n(n + 2)log-R - r

+nlog+V(R,f),

O
Proof: Fix such an n and R , and let such anfbe given. Let h be a holomorphic logarithm off in D(0, 1) and apply the Cauchy-Schwarz formula (5.1 8) to h in D(0, R ) : h(z) =

& IO2'Re h(Refe)[Rei* - z

+ i Im h(0)

Omitted Values and Normal Families

418

Apply 2.14 to each of these latter integrals and amalgamate the results to conclude that

Let F = f'p Repeated differentiations under the integral in (1) (i.e., appeals to 2.14) give

o'+ l)! 2,, Joa'

(2) F"'(z) =

2Reie logIf(Re'e)l (Relo - z)'+I dd, I z I < R , j = O , l ,

....

It follows that

- 2RO' (R - r ) ' + 2

(3)

Thus, since 0 < R (4) IF(')(re't)l

V(R,J?, t E R, r < R, j = 0, 1,. . ..

- r < R < 1,

I2(n

+ l)! (R - r)-n-aV(R,f),t E R, r < R, j = 0, 1,. . ., n.

Use the fact that log+ is a non-decreasing function and log+@!x) Ilog(n!)

+ log+(x)

(12.1(ii))

together with the last lemma and (4), to see that

I

fo Ilog(n!) + log+[2(n + l)! (R - r)-n-aV(R,f)J",

log+/

0 2. Miranda's Theorem

419

t E R, r < R. Then use 12.1(ii) again to get

1 + l)!) + (n + 2) log Fr =

(5) t E R,

log(n! (2(n + I)!)")

+ n(n + 2) log R- r + nlog+ V(R,f),

r < R. Christen this unpleasant constant A, and integrate in ( 5 ) to get L(r,

7)

I A,

1 + n(n + 2) log R - r + n log+ V ( R , f ) ,

r < R,

as desired.

Lemma 12.10 (Bureau's Bootstrap Lemma) Let 0 I a < b I 1 and f be a non-negative, non-decreasing function on (a, b). Suppose that for some positive constants A and B, f satisfies (1)

f ( r ) 5 A log+f ( R ) + A log R-r + B V a < r < R < b .

If B 2 2A(A + 2), then f (2)

also satisfies

+2BVa
Hence, regardless of the size of B, f satisfies (3)

f ( r > 5 2A log Tr + 28

+ 4A(A + 2)

Va < r < b.

Proof: Suppose that (4)

B 2 2A(A

+ 2)

and that nevertheless (2) fails : (5)

f(ro)

'2A 1%

+ 2B

for some a < ro < 6 .

Then ( 5 ) also holds with b replaced by b' < b, i f b' is sufficiently close to b. Moreover, f is bounded on (a, b'), by f(b'). Therefore without loss of generality we can suppose, in trying to deduce a contradiction from (l), (4) and ( 5 ) , that f i s bounded on (a, b). Set ro rl=-=

+b

2

ro

b - ro +2

420

Omitted Values and Normal Families

and apply (1) with r = ro, R = rl to get

f(ro) I A log+f(rl) log+f(r1) 2

1

+ A log +B b - ro b-r, + log 2 -B

- p O )

1

9

whence, recalling 12.1(iv),

m

1

2 zf(r0) + log-

> 2log= log-

B >2

(6)'

b

b-ro 2

- A-B

- A- ro + 7 + log 2 1

2B

2 b-ro+l

- log4

b-ro

B

by (5)

- log4

(since b

- ro < b - u 5

1).

Now from (4) B 2 2A(A

+ 2) = 2A' + 4A

B - 4A 2 2Aa

(7)

; Bi- log4 >

:-

4 2 2A

and also from (4)

+ 2) 2 2A(A + log4) B' - 2AB(l0g 4 + A) + A' log' 4 2 0 (B - A log 4)' 2 2A'B B 2 2A(A

(8)

(-ilog 4)'

2 28.

Now multiply inequality (6) by (6)', keeping (7) and (8) in mind:

2 2A log -+ 2 B

- ro = 2A log -+ 2B. b - rl b

(9)

9 2. Miranda's Theorem

421

From the ( 5 ) to (9) implication it follows that if we define r, =

+ b, -

r,-l

n = l , 2 ,...,

then

f(r,,) > 2A log -+ 2B. b - r,,

Since b

- r,, = (b - ro)/2", this gives

f ( r ) > 2A log- 2" b - ro

+ 2B

and this makes f unbounded, a contradiction. Thus (2) is established. For (3), just replace B by B quote (2) for this new constant.

+ 2A(A + 2) in (1)

and

Lemma 12.11 For each non-negative integer n there exists a constant B,, such that every f which is holomorphic and zero-free in D(0, 1) and satisfies (i)

If(")(z)/1 1 Vz E D(0, 1)

also satitjes

Proof: (1)

We have from 12.2 and 12.l(v) V(r,f) = L(r,f)

+ L( +)

and from 12.3 and 12.1(v)

and therefore from 12.9

Omitted Values and Normal Families

422

holds for 0 < r < R < 1 . Use (3) in (2) to get

(4)

+ log+If(0)l + 2n(n + 2) log Fr+ 2n log+ v ( R , ~ ,

V(r,nI 2 4

holding for 0 < r < R < 1. Now by (1) and 12.5, V ( r , f ) is a non-decreasing function of r, so we can apply Bureau's lemma 12.10 to conclude that

+ 2 log+If(0)l + 4(2n(n + 2))(2n(n + 2) + 2) 1 + 4n(n + 2) log 1-r

V ( r , f ) I 4A,

= 2 lOg+lf(O)l

(5)

+ an + bn log

1

-

9

O
(with the obvious definitions of a, and b,). Next apply 12.4(ii) with R = ( 1 + r ) / 2 to see that

Apply ( 5 ) and (6) together to get (7)

log M ( r , f ) I

& [8 log+If(0)l + 4a, + 4b,

Choose for B, any number which satisfies B, 2 8b,

8an and B, 2 log 2

Then B, log -2 fB, log 2 1-r

2 + fB, log 1-r

+

2 4 ~ , 4b, log

2 1 -r'

O s r < l

and (7) gives

(8)

1% M ( r , f ) I

[8 log+lf(o)1+ B n l o g r r

]

9

O
(By continuity the inequality (8) holds as well at r = 0.)

Lemma 12.12 For each non-negative integer n there exists a constant C, such that every f which is holomorphic and zero-free in NO, 1 ) and satisfies (i)

If(")(z)l I 1 Vz E D(0, 1 )

g 2. Miranda’s Theorem

also satisfies (ii)

log M ( r , f ) I

423

[16log+lf(O)l + C , l o g r r ] 1-r

VOIr<1.

Proof: We go by induction on n. If n = 0, (i) says that C, = 0 will work in (ii). Suppose the result true for 0 I n < N and consider n = N. Let such a function and an r E [0, 1) be given and set r ’ = (1 + r)/2. Consider two cases : Case I: There exists z’

(1)

E D(0, r‘) such that

If”-”(z‘)l I 1.

Then for all z E D ( 0 , l ) If‘”-”(z)l I If“-”(z’)l I

+

< 3.

I

+ If(”-”(z) - f(N-l)(z‘)I

I

(N)

5 1

+ Iz - z’I

by (i)

Apply the induction hypothesis to the function +f:

whence, since log+I+f(O)l 5 log+If(O)l,

+

log M ( r , f ) = log M(r, if) log 3 1 I1 - r [16log+lf(O)1

+ C,-,log- 1 - r ] + log3

Case ZZ: (3)

lf”-”(z)I

> 1 vz E D(0, r’).

In this case the function F(z) = f(rfz)/(rf)N-l(121 < 1) satisfies the hypotheses of the last lemma with n = N - 1 and so we have (4)

1

1 * r R [8 log+IF(0)l + BN- log 1-R’

log M(R, F ) I

We have for 0 I R < I

01R<1.

Omitted Values and Normal Families

424

In particular then, =

log M(r/r', I;)

and (4) with R = r/r' gives

1%

W,f1 5

(- f)

log M r, (;

l+r 1-r

I

')!N

1

[8 log+lf(0)1

+ (N - 1)s l o g 2F r

by definition of r' and by 12.1(ii), 1-r I

1-r

16log+lf(O)J

, + 16(N - 1)log-I +2 r + ~ B N -logI -4 r 1

[16 log+[f(0)l + (16(N - 1) + 2B~-1)log 2 16 lOg+(f(O)14- (16(N - 1) -I-4BN-1)

From (2) and ( 5 ) we are lead to define

{

CN= max CN-1

log 16(N - 1) + 4BN-1 +log2'

and thereby have (ii) valid for n = N.

Lemma 12.13 For each non-negative integer n there exists a constant K,, such that every f which is holomorphic and zero-free in D(0, 1) and satisfies

also satisfies

8 2. Miranda's Theorem

425

Proof: Let such a function and an r E [0, 1) be given and set r' Consider two cases as before:

= (1

+ r)/2.

Case I: There exists z' E D(0, r ' ) such that

(1)

If("'(z')l

I

1.

Let F be a holomorphic logarithm for f(") - 1 in D(0, 1):

(2)

f(")- 1 = ep

Re F(z') = log1f(")(z') (3) As in the last proof

- 1I (1)I log 2.

+

ReF(z) IReF(z')

+ Iz' - zI < log2 + 2

Ilog 2

(4) It follows that (5)

If(n)l

Let c = 1 (6)

I1

+ If(*) - 11

+ 2ea and form

1

by (i) vz E D(0, 1).

+ PF

5 1 + 2ea.

(4)

z E D(0, 1).

F(z) =

Because of (5) we may apply the last lemma to P:

(7)

1 r R [16 log+1~(0>1+ c n log rR ] VOSR
log M(R, PII

We have for 0 I R < 1

In particular then, f and (7) with R

= r/r'

)

=

log M ( r / r

1,

F)

gives

) log c log ~ ( r , fI Ilogc

+ log M

+

since c > 1.

= log c

rlr ' [161og+(%l

+ log M(r/r',P )

+ Cnlog-1 - r / r '

Omitted Values and Normal Families

426

Appraisals like those in the proof of inequality (5) of the last lemma then lead us to log W , f 5 ) log c

+ rr [3210g+lf(0)1 + (32n + 4Cn) 10gT-T;

Case ZZ: (9) lf(n)(z)I > 1 Vz E D(0, r'). This is exactly case I1 of the last proof and so, as was deduced there, log M(r,f) 5

(10)

Sr[

16 log+ If(0)l

+ (16n + 4 4 ) log

From (8) and (10)it appears that the choice

+ 4Bn,32n + 4Cn + log 2

16n will validate (ii).

Theorem 12.14 (MIRANDA[1935]) For each non-negatiue integer n there exists a constant Mn such that every f which is holomorphic in D(0, 1) and for which f and f("' - 1 are zero-jiree satisfies

Proof: Let such a function f and an r E [0, 1 ) be given and set r'

+

= (1 r)/2. Three cases arise, of which the first two are trivial reductions to previous lemmas :

Case I: (1) If(")(z)I4 2 V z E D(0, r'). Then apply 12.12 to the function F(z) = f(r'z)/2(r'y and, via the by-nowfamiliar routine, translate the resulting inequality about F into one about f. There results log

w,f)

c log 2 < log2 5

1-r

Fr[ +' +

16 log+lf(O)1

1 -r

+ 16n log( 1 : r ) + cnlog- l - r 2(1 + +

2 [32 log+If(0)l + 32n log - 2Cn log l -4r 1 l+r

[32 log+If(0)l + (32n + 2Cn + 1 ) log 2 + 2Cn log 1 -' r 1

"I

4 2. Miranda's Theorem

427

Case 11:

Then for the function F(z) = f(r'z)/(r')"

we

P ( z ) = f(n)(r'Z),

u ience

and so 12.13 is applicable to F. After the canonical maneuvers there results 2 [64 log+lf(0)l + 64n log -+ 2K, log rr 14 -r 1 l + r

log M ( r , f ) 5

<

(4)

1-r

[64 log+If(0)l

Case III:

If(")(z')I

(5)

and

1.

+ (64n + 4Kn) log 5 -

> 2 for some z' E D(0, r')

Ij & ~1 ~ ~ ) l

> 1 for some z" E D(0, r').

(6)

From (6) we have in particular that f ( " + l )is not identically 0 in D(0, 1) and so, as noted before (penultimate paragraph of the proof of 12.4), f ( " + I ) has a zero on C(0, p) for at most countably many p E [0, 1). In what follows consider P

(7)

'2" +

such that 0 $f(""(C(0, p)). +

Then on C(0, p) we may write 1 f(n) f( n ) - 1 f(n+l) =---.-

7

f

f(n+l)

f

and apply 12.l(ii) and (iii) to see that

As noted in the opening lines of the proof of 12.11, V P , f ) = 2L(P,j)

and so

+ h3lf(O)l,

Omitted Values and Normal Families

428

Apply 12.4(iii) to F = f(n+l)/(f(n)

- I),

z = z" and R = p :

(9) Apply 12.9 with the n there equal 1 and thefthere equal f(n) - 1 : (lO)L(p,

fG) + 5

Al

3log-

+ log+V(R',f")- l),

1 R - P

p

< R' < 1 .

Similarly appraise the terms L(p,f(n)/f) and L(p, f("+l)/f) in (8) via 12.9 and use (9) and (10) to get V h f ) 5 lOg(f(0)l

+ (log4 + 2 4 + W+3

1 + 2n(n + 2) log + 2(n + l)(n + 3) log R- P R-P + 2n log+ V(R,f ) + 2(n + 1) log+ V ( R , f )

128

-

[A1

3lOg-

R'

- p + log+ V(R',fCn) -

holding for all p c R' c R < 1 . Take R' = + ( p V(p,f) 5 loglf(O)l

+ R) and get

+ (log4 + 2An + 2An+d

1 + 2(n(n + 2) + (n + l)(n + 3)) log R-P 128A1 2 + 2(2n + 1 ) log+ V(R,f ) + + -log384 (1 - r)a (1 - r)a R-p

+-(1 128 - r)a log+ V(R', - l), f(")

holding for all p < R < 1. Set (1 1)

+ 2 4 + 2An+1+ 2n(n + 2) + 2(n + l)(n + 3) + 128A1 + 384

H,,= log 4

0 2. Miranda's Theorem

429

and have then a fortiori (12)

V ( P , ~ ) logIf(0)l Hn +-

Hn

+ Hn + -log(1 - r)a [log+ V(R,f)

R

2

-p

+ log+ V(R',ftn)

- l)],

+

holding for all p < R < 1, R' = i ( p R). As in the derivation of (9), use the fact If(")(z')- 1I > 1 (from (5)) and 12.qiii) to deduce that

( fen)'- 1)

R' (R'

L R',-

+ I4 - Iz'])

aL(R',f'n) - 1)

1 1

I-

+ L(R',f) + log2 ,

by 12.l(ii)

+ V(R,f ) + log 2 ,

since R' c R.

Now apply log to (1 3), again making use of 12.I : +

log+ V(R',f'")- 1)

I log

l7 + log+ L( R', 7 )+ log+ V(R,f ) + log+ log 2 + log 3 (R - PY ftn)

F)+

17 < log - L(R', log+ V(R,f) + log 3, ( R - PIa using the obvious facts log+ x Ix and log 2 < 1.

+

Omitted Values and Normal Families

430

An application of 12.9 then gives

1 l7 A , n(n 2) log R - R' ( R - PIa n log+ V ( R , f ) log+ V ( R , f ) log 3 51 2 2 = A, log7 2logn(n 2) log R-P R-P (n l)log+ V ( R , f ) .

log+ V(R',f(")- 1) 5 log

-+

+

+

+ +

We set (15)

+

HA = A,,

+ +

+

+

-

+ +

+ l o g y + (n + l)(n + 2)

and have (Ifortiori from (1 4) (16)

log+ V(R',f(")- 1) IHk

If we insert (16) into (12),

+ Hn(H' (1 - r)l +

+ H;lOg- R - P + Hilog+ V ( R , f ) .

log+ V ( R , f )

I ) log+ V(R,f) + Hn(H' (1 - r)a +

+ HAHA (1 - r)a +

log+ V(R,j-).

This inequality is valid for a dense set of p E ((1 + r')/2, 1) and for each such p it is valid for all p < R < 1. But an easy uniform continuity argument on the integrand defining V ( p , f ) shows that it is a continuous function of p and so it follows that (17) is valid for all p ~ ( ( 1 r')/2, I ) and all p c R c 1. Since (12.5), V ( p , f ) is a non-decreasing function of p, we are in a position to apply Bureau's lemma 12.10. We get

+

4- I) 1 log + 2H,(H,: (1 - r)a 1 - p'

l t 2r ' < p < l .

6 2. Miranda’s Theorem

43 1

Let

(19)

L, = 4Hn(H1:

+ 1)

and have a fortiori from (18)

1 log +-(1 Ln - r)a 1 - p’

2

+

r’ < p < 1.

An eligible p is (21)

3+r’ 4

p = --

We have p

7 + r

-- 8 ’ - r = i(1 - r ) > (1 - r)/2 so we may apply 12.4(ii) to see that

Use (21) in (20) and insert the result into (22):

(since log x 5 x for all x > 0) I

8 48L’ (since L, 2 I). log+If(O)] + 1 - r (1 - r ) s

From (2), (4) and (23) we see that we may take for M , the number

Mn = max(64n

+ SC, + 2, 128n + 8Kn,48Li).

Exercise 12.15 (Schottky) When n Miranda’s theorem, namely log M ( r , f ) I

=

0, a stronger conclusion is actually valid in

[I2 log+If(0)l + M ; log 1-r

O s r c l

Omitted Values and Normal Families

432

for some constant MA. Prove this by refining the argument for case III above, notably by citing 12.4(iii) (with F = f'/Cf - 1)) to assert that

iff'(0) # 0, and using (9)' instead of (9) in the rest of the proof. Similarly, get a simple estimate in (13) free of the factor 1/(R - P ) ~ .

03

Immediate Applications of Miranda

Corollary 12.16 If F is a zero-jiee entire function and for some non-negative integer n the range of F(")does not contain all non-zero complex numbers, then F is a (non-zero) constant. Proof: If C\{O} F(")(C), then by scaling we may suppose that 1 4 P(C). Consider R 2 3. Set f ( z ) = F(2Rz)/(2R)". Then f is holomorphic and'zerofree in D(0, 1) and f(")(z)= F(")(2Rz), so f ( " ) - 1 is also zero-free. Take r = 3 in Miranda's theorem to conclude that

+

I 128 log+lF(O)J+ 32Mn, (1) since log+ is non-decreasing and 2R 2 1. Let B denote the number on the right side of (1). Then (1) says that sup If(z)l s eB, that is, Is1 s u a

(2)

sup IF(z)l I(2"eB)Rn,

121 % R

holding for all sufficientlylarge R. It follows from this and 6.33 that F is a polynomial of degree not greater than n. Since F has no zeros, it must in fact be constant (4.49). Corollary 12.17 (Picard's Little Theorem) At most one complex number is absent from the range of a non-constant entirefinetion. (Cf. 12.28.) At most two complex numbers are absent from the range of a non-constant function which is meromorphic in the whole plafie. Proof: I f f is entire and a, b $f(C) where a # b, then the entire function F = f - a is zero-free and 0 # b - a $ F(C), so by 12.16 F is constant. If G is rneromorphic in C and a, b, c are distinct complex numbers not in the range of G, then 1/(G - a) extends to an entire function which omits l / ( b - a) and l/(c - a). Remark: The non-constant functions ez and 1/(1 - 8) are examples where one or two values, respectively, are omitted.

6 3. Immediate Applications of Miranda

43 3

Corollary 12.18 (P6lya-Saxer) I f f is an entire function and f a $ ’-f”is zero-free, for some constants a and b. then f ( z ) = eaz+b Proof: f never vanishes, hence is of the form f = eg for some entire g and g’ = y . e - 0 never vanishes, hence has the form g’ = eh for some entire h. Then f’ = g’P = P + h ,so g’ + h’ = f”.e-(g+h)never vanishes, that is, eh + h’ never vanishes. Therefore eh(l + h’e-h) = eh(l - (e-h)’) never vanishes. It follows that (e-h)‘ does not take the value 1. 12.16 then implies that e-h must be constant. Since g’ = eh, it follows that g is linear and since f = P, the desired conclusion about the form off is obtained.

Corollary 12.19 a fixed point.

Iff

is entire and not of the form f ( z ) = z

+ b, then f

0

f has

Proof: Supposef f has no fixed point. Then of course f has none either and 0

therefore we may form

an entire function. We have g is zero-free since f 0 f has no fixed points. Also g(z) # 1, else f ( f ( z ) ) - z = f ( z ) - z and f ( z ) is a fixed point for$ By Picard’s Little Theorem g is then constant, say c, where c # 0, 1. (1)

f ( f ( z ) ) - z = c ( f ( z ) - z ) v z E c.

Differentiate ( 1 ) to get (2)

f’(z)[f’(f(z)) - c] = 1 - c v z E c.

It follows that f ’(f ( z ) ) # c for any z and thatf’(z) # 0 for any z. A fortiori from the latter, f ’ ( f ( z ) ) # 0 for any z. One more application of Picard’s Little Theorem shows thatf’ f is constant. Then (2) shows that f ’ is constant, say a. Setting b = f(O), we thus havef(z) = az + b. This function has a fixed point (namely -b/(a - 1)) unless a = 1. 0

Theorem 12.20 Let f and g be entire functions and for some positive integer n satisfy the identity f ” + g ” = 1. (i) (ii)

Zfn = 2, then there is an entire function h such that f Zfn > 2, then f and g are each constant.

=

cos 0 h, g

=

sin 0 h.

Omitted Values and Normal Families

434

Proof: (i) When n

Cf+

=2

the identity is

i g u - ig) = f a

+ ga = 1.

Thus the entire function f + ig has no zeros and consequently is of the form elh for some entire h. It follows then from (1) that

Therefore

1 1 g = Zcf+ ig) - 2 ( f- ig) =

elh

- e-lh 2i

=

sin h. 0

Let u be a primitive nth root of - 1, say u = etn/".Then u, u3, u6,. . ., are all the nth roots of - 1 and so

u2n-l

zn + 1 = (Z - U)(Z

- u3)(z - u6).

- .(z - uan-l),

VZEC

If u, w E C and w # 0, put z = u/w and multiply by wn to get u"

+ w" = (u - uw)(u - u3w)(u - U")'

*

- ua"-'w).

'(0

Of course this last equality is valid for w = 0 also and so 1 = f"

+ g" = (f - ug)Cf - u3g)(f - u").

*

*(f

- 24"-'g).

Thus each factor on the right is an entire function without zeros and there are at least three such factors since n 2 3. Consequently there exist entire functions h,,, . . .,h,- I such that

f - uak-'g = ehk-1, k

= 1,2,

3 , . . ., n.

Subtract equation (3.2) from equation (3.1) and equation (3.3) from equation (3.2) and get (u3 - u)g = eho - 8

1

(v6 - u3)g = e"l - @a. Now u # 0 and u2 = eaniln# k 1 (since n > 2) and so (4), and (5) give

- &) = ehi - @a u2eho + eha = (ua + 1)e"l u2(@0

8 3. Immediate Applications of Miranda

435

where v is any square root of the complex number u2 + 1. Applying (i) to equation (6) provides an entire function h such that

It follows from (7) that h is constant. For if not, then by Picard's Little Theorem there exists a z E C such that either h(z) = ~ / or 2 h(z) = - 4 2 . In either case cos h(z) = 0 and equation (7) is violated, for an exponential is never 0. From the constancy of h follows that of eho-hl,by (7). Let, say, = a.

$o-hi

Then from (4) eho - ehi = ehi eho-hi - 1 = ehi a - 1 u(u2 - 1) - 1) U(Ua - 1)

= u(ua

= Aehi,

where A =

a- 1 u(u2

- 1)'

while from equation (3.2)

f = u3g + $1 =

= u3Aeh1

+ ehl

Behi, where B = u3A

+ 1.

From (lo), (1 1) and the basic identity 1 =f"

+ g" = (A" + Bn)enhi,

it follows that A" + B" # 0 and then that enhiis constant. Thus by (lo), (ll),f" and g" are each constant. The range of the holomorphic function fthus lies in the finite set composed of nth roots of the constantfn. By the Open Map Theorem (5.77)fis therefore constant. Similarly for g.

Exercise 12.21 (Cf. 7.25) Let .% denote the set of all holomorphicfunctions f in D = D(0, 1) withf(0) = O,f'(O) = 1 and 0 $f(D\{O}). Show that there exists an a > 0 such that D(0, a) c f ( D )for every f E Yo.

Hints: Suppose contrariwise that there exist u, E C\{O} with an +0 and fn E 9, such that a, $f,(D). Then 1 - fn/an is zero-free and so there exists for it a holomorphic logarithm L , in D , which may be normalized by L,(O) = 0. The numbers f 27ri are then absent from the range of each L, and so by 12.14 {L,} is uniformly bounded in D ( 0 , t ) . From this and 5.32 follows the boundedness of the set {( 1 - fn/an)'(0)} = { - 1/a,}.

436

Omitted Values and Normal Families

04

Normal Families and Julia’s Extension of Picard’s Great Theorem

Definition 12.22 A set 96 of functions holomorphic in an open subset V of C is called normal (in V) if every sequence in 96 contains a subsequence which either converges locally uniformly in V or diverges to 03 locally uniformly in V. Lemma 12.23 (MONTEL [1916], p. 227) Let R be a region, .Fa family of functions holomorphic in R with the property that about each point z E R there is an open disk D, in which 96 is normal. Then .@ is normal in R.

Proof: Let {f,} be a sequence in S.Since !J is a countable union of compacta (e.g., the sets {z E C: IzI In and d(z, C\R) 2 l/n}), it is easy to see that countably many of the disks D, cover R. Since 96 is normal in each of these, a diagonal argument produces a single subsequence {f,,} of {f,} with the property that every point z of !J lies in some open disk on which {fn,}either converges uniformly or diverges uniformly to 00. If the two classes of points z are denoted by C and D respectively, then by their very definitions each is open, C n D = 0 and C u D = R. It follows from the connectedness of R that either C = R or D = a. A simple covering compactness argument shows that in the first case {f,,} converges uniformly on each compact subset of R and in the second case {fn,} diverges uniformly to 03 on each compact subset of R. Theorem 12.24 (MONTEL [I9113) If U is an open subset of C,S c H(U)is a normal family and for some a E U the set S ( a ) = {f ( a ) :f E 96) is bounded, then 96 is locally uniformly bounded on U. Proof: Suppose contrariwise that .@ is not uniformly bounded on some compact K c U.So there exist f , E S with

(*I

suPIfn(K)I > n.

A subsequence {f,,} of {A}either converges locally uniformly on U or diverges locally uniformly to 03 on U.Since F(a) is bounded, the sequence {h,(a)} is bounded. Hence the second part of the alternative is untenable, the first part prevails and (*) is contradicted. Theorem 12.25 Let R be a region, N a non-negative integer, a, b E C , b # 0 ; and a # b is N = 0. Let 96 be a family of holomorphic functions in R such that for each f E S,a 4 f ( R ) and b 4 f (N)(R).Then S is a normal family. Proof: For each f

E 96 define

$ 4 . Normal Families and Julia's Extension of Picard's Great Theorem

437

and let $ = { f l f S}. ~ Manifestly 9t is normal if and only if $ is. Moreover for each f~ $ we have 0 $f(!2) and 1 $f("(R). We are therefore reduced to considering the case a = 0, b = 1. We first establish that, for each R > 0 there is a constant MN(R)such that (1) D(w, 2R) c !2 * lOglf(Z)l I128 lOg+If(w)l

+ MN(R) V z E D(w, R ) , f € 9.

Indeed, given that D(w, 2R) c a,it is possible to define for each f E 9 a function F on D(0, 1) by F(z) = f(2Rz w ) / ( ~ R )This ~ . function meets the hypotheses of Miranda's theorem and so, taking r = f; in that theorem,

+

Sup lOglF(Z)l 2 2[6410g+IF(O)I

16M~],

IZI L l l 2

that is,

loglf(z)l I 128 log+If(w)l We set MN(R)= 32MN from (1) that (2)

+ 32MN + 127110g(2R)~I Vz E D(w,R).

+ 127Nllog(2R)I and

D(w, 2R) c R s sup log1f(z)l ZED(W.R)

s

have (1). It follows at once

128 log+ If(w)l

+ MJR)

Vf

+ MN(R)

Vf

E 9t

and also that (3)

D(z, 3R) c !2

loglf(z)l I 128 inf

weD(z.R)

log+If(w)l

E 9,

since D(z, 3R) c !2 implies D(w, 2R) c Q for every w E D(z, R). Given any point u E R, choose R = R(u) > 0 so that D(u, 3R) c R. We will show that S is normal in D(u, R). Then we can cite 12.23 to assert that 9 is normal in R,as desired. To prove that 9 is normal in D(u, R), let a sequence {f,}in S be given. We want to show that either there is a subsequence of {f,} locally uniformly convergent on D(u, R) or there is a subsequence of {fn} divergent to 00 uniformly on D(u, R). In case {f,(u)} is bounded, take w = u in (2) and learn that {f,}is uniformly bounded in D(u, R). Then cite 7.7 to come up with a subsequence of {f,}locally uniformly convergent on D(u, R). In case {f,(u)} is unbounded, there exist 1 I n, < n2 such that = co. Then (3) with z = u shows that limk+mlfnk(w)l= 00 1irnk~,.,,~fnk(u)~ uniformly for w E D(u, R). <

.

a

-

Lemma 12.26 Let U = { z E @: 0 < IzI < I } and let f be holomorphic in U and have an essential singularity at 0. Defne fn(z) = f(2-"z) ( z E U ) . Then {f,}is not a normal family.

Omitted Values and Normal Families

438

Proof: (MONTEL[1927], pp. 79,80) According to Casorati-Weierstrass (1 1.6)

there are points z arbitrarily near 0 on which f takes arbitrarily small values, that is, there exist z k such that (1)

1/4 >

1 ~ 1 1> 1~11> *

- *+O

and

Choose integers nk 2 1 such that 2-nk-a 5

Izkl

< 2-S-l

for each k. Then {nk} is a non-decreasing and unbounded sequence. We have (3)

wk

= 2"Wk E

k = 1, 2, . . .

z($,+),

and (4)

fnk(Wk)

= f('k)

as

*'

Now let us suppose that (A}is a normal family and derive a contradiction. The sequence {fnk}in the normal family {f n }contains a subsequence {fnk,)which is either uniformly convergent on each compact subset of U or is uniformly divergent to a on each compact subset of U. On the compact subset z(*,+), however, {fnJcannot diverge uniformly to m, as (3) and (4) show. So the first alternative holds. In particular, {fnJ is uniformly bounded on the compact subset &, +), say by M. Given z E D(0,2-na-1)\{0},there exists j 2 1 such ~ ( z ( < 2-"*1-'. Then 2 b z E & 3) and so that 2 - " * 1 - 5

If(')[

= Ifk,(2"k'z)I

M*

This shows thatfis bounded by M in D(0, 2-"*1-1)\{0}, making 0 a removable singularity (by 5.41), contrary to hypothesis. Theorem 12.27 (Julia) If the holomorphicfunction f has an essential singularity at a, then there exist real number do and complex number b such that for every (suflciently small) e > 0 (I)

C\{b} C f { a + ref8:10

- 0,l

< e, 0 < r <

8).

Remark: Any such ei8ois called a Julia direction and the ray {a is called a Julia line (for the functionf).

+ re'%: r E }'W

Proof: Without loss of generality we may suppose that a = 0 and that f is holomorphic in U = {z E @: 0 < Izl < I}. Let r,, = 2-" andf, be as defined in

4 4. Normal Families and Julia's Extension of Picard's Great Theorem

439

12.26. By 12.23 then there is a point zo E U such that {fn} is not normal in any neighborhood of zo. For every pair of positive integers n and m set

We will show that for any two distinct complex numbers b and c and any m, there exist infinitely many n for which either b ef(Dn,,) or c € f ( D n , , ) .[For a set S not known to be in the domain offwe write simplyf(S) for thefimage of those points in S which are in the domain off. If there are none,f(S) means the void set.] Forf(D,,,) = fn(D,), so if the assertion is false there is an m and an N such that all the functions fn for n 2 N omit the distinct values b and c on D,. But then by 12.25, {fn}FxN(and so obviously also {fn}E1) is normal in the neighborhood D , of zo, contrary to the definition of z,,. It follows that for each m there is a complex number b, such that k m l n=k

Pick Oo so that zo = lzOlefeoand let (4)

S

=

S(e) = {refe:16 - Ool <

E

> 0 be given and consider the sector

e,

0 c r < e}.

It is open and if k is such that rk < e / l z o J then , rnzO= rnlzoleieobelongs to S for all n 2 k (because for such n, rnlzOl IrklzOl < e). We may then choose m = m(e) large enough that Dk,, =

{

ZE

@:

IZ

1 - rkZOl < r

IZ

I}

m k o

c

s.

Then for any n 2 k rn Dn,m = - Dk,m rk

s7

since S is clearly closed under multiplication by positive scalars less than I . Thus

u m

(5)

n= k

Dn,m c S-

From (3) and (5) follows

(6)

Q=\{brn} c f ( S ) .

The assertion of the theorem follows from this. For if there are two drfterent complex numbers b' and b" such that 6' $f(S(&'))for some 8' > 0 and 6" $f(S(e"))

Omitted Values and Normal Families

440

for some e" > 0, then for e = min{e', e"}, both the complex numbers b' and 6" would lie outside f(S(e)), contradicting (6). Exercise 12.28 Deduce from 12.27 the following stronger form of Picard's Little Theorem: a non-polynomial entire function assumes infinitely often every complex value with at most one exception-Rcm [18801. Hints: If a # b and a, b are each assumed only finitely many times by the entire function F, then neither is assumed outside D(0, I/r) for sufficiently small positive r. Consequently the functionf(z) = F ( l / z ) , 0 < IzI c r, does not have an essential singularity at 0, i.e., its Laurent series contains only finitely many negative powers of z. Evidently then, F is a polynomial. Exercise 12.29 Let SZ be a region, cf,} a sequence of univalent holomorphic functions in a. The object of (i)-(iv) below is to show that there exists a point a E R and a subsequence {hk}of {f,} which is normal in SZ\{a}; (v) and (vi) are applications of this fact. (i) There is a point a E SZ and a subsequence { f , } of {f,} such that in each compact K c SZ\{a},for only finitely many j does the function A, have a zero.

Hints: If the set of subsequential limits of {f ; '(0)) meets SZ, let a be any point in the intersection. Otherwise, let a be any point whatsoever in SZ. Let {K,}be a compact exhaustion of R\{a} as furnished by 1.31. For eachj let N, be a neighborhood of a in @a, llj) n a which is disjoint from K,.In the first case there exists a subsequence {f",}such that fn;l(O)E N, for each j . In the second case no set { f ~ l ( o ) }(k~ = ~ kI , 2, 3, . . .) lies wholly in any Kfand we can choose the n, successively such that I I n, < n, < . - andfG'(0) 4 K,.Thus in either case K, n {f;;;l(O)}S = 0 , for each j . Since any compact subset K of Q\{a} lies in some K,,the claim (i) is established. (ii) There is a point b E Q\{a} and a subsequence {hk}of {fn,} such that in any compact K c SZ\{a, b},for only finitely many k does the function hk lake either of the values 0 or 1. +

Hints: Apply the argument of part (i) to the sequence (1 functions in the open set n\{a}. (iii) The sequence {hk}of (ii) is normal in a\{a, b}.

- f,,}

of univalent

Hints: By (ii) and 12.25 {hk} is normal in every open disk D such that c SZ\{a, b}. Since the latter is connected (1.24), it follows from 12.23 that {h,} is normal therein. (iv) The sequence {hk}of (iii) is normal in a\{a}. Hints: Since (hk} is normal in B\{a, b}, any subsequence of {hk} contains a further subsequence {g,}such that either (I) : {g,}converges locally uniformly in

4 5. Sectorial Limit Theorems

441

Q\(a, b} or (2): { J g f ldiverges } to m locally uniformly in Q\{a, b}. If (1) prevails, then { g , } converges uniformly on the boundaries of two disjoint closed disks Do, Db centered at a and b and lying in Q (since such circles are compact subsets of Q\{a, b}). Then by the Maximum Modulus Principle {gJ converges uniformly in Do and D,. Thus ( I ) implies (I)': {gi}converges locally uniformly throughout Q. If (2) prevails, select r > 0 so that b ( b , r ) c O\{a}. Then Ig,l +. co uniformly on C(b, r ) . Bearing in mind the definition of the fn, from (i) and the fact that {g,}is a subsequence of {fn,}, we know that for only finitely many I does g, have a zero in B(b, r ) , so { 1 /g,} is a sequence of holomorphic functions in D(b, I ) . From 11/g,l +0 uniformly on C(b, r ) and the Maximum Modulus Principle it follows that I l/g,l +.0 uniformly in D(b, r ) , so lg,l --f co uniformly in D(b, r ) . Thus (2) implies (2)': {lgfl}diverges to 00 locally uniformly throughout Q\{a}. (Extension of7.29(i)) Show that if (v)

(*)

{fn(c),fn(d): n E N} is bounded for two distinct points c, d E Q, then {fn) is locally unifbrmly bounded in

a.

Hints: Consider any subsequence of {fn}. Apply (iv) to this subsequence to come up with a point a E Q and a further subsequence { g , } of the given subsequence, such that either (I): {g,} converges locally uniformly in Q\{a} or (2): {lg,l} diverges to co in Q\{a}. Since one of the points c, d must be in Q\{a}, alternative (2) cannot prevail because of (*). Now argue as in (iv) that { g , } converges uniformly in some disk centered at a, hence locally uniformly throughout Q. We have shown that {fn} is normal in a.Now conclude by citing 12.24.

(vi)

Prove the following extension of 7.11 (due to MONTEL[1925], p. 253): Iflimn+mfn(c),limn+mfn(d) exist in C and coincide, equal L, say,for two distinct points c, d E Q, then {fn} converges to L locally uniformly in Q.

Hints: Use (v) above, 7.1 1 and 7.8.

85

Sectorial Limit Theorems MONTEL[I9121 used normal families in a simple but ingenious way to investigate boundary behavior of holomorphic functions in angular domains. (Later he used these ideas to treat the boundary problem for conformal maps which we analyzed in Chapter IX.) We have already explored this theme, on the boundary itself in 5.16 and internally in 5.56. Most of the theorems of this section look at internal approach. The bounded version of 12.30 is, of course, contained in 5.56. It is interesting to note that, in spite of the diversity of proof techniques involved, the bounded 12.30 plus 5. I6 is equivalent to 12.3 I . Theorem 12.30 Let ct < y < t9, S = {reis:r > 0, ct < 0 < 8) and let f be holomorphic in S and for some complex c satisfy lim,-m f(reiy)= c. Suppose

Omitted Values and Normal Families

442

that there are two distinct complex numbers absent from the range o f f . (For example, f might be bounded.) Then limr+mf(refe)= c for every 8 E (a,/?)and uniformly in any compact subinterval.

Proof: Note that S is an open, convex (hence connected) subset of C. Define for each positive integer n functionsfn in S by (1)

E S.

fn(z) = f ( n z )

Notice that (2)

lim fn(refY)= c Vr > 0.

n+ m

In particular, the set {fn(efy)}is bounded. Since also the two distinct numbers absent from the range off are both absent from the range of each fn, it follows from 12.24 that {fn} is uniformly bounded on each compact subset of S. But then from (2) and the Vitali-Porter theorem 7.6 we know that fn converges to a holomorphic function fo uniformly on each compact subset of S. On the ray (0, m)efy we have fo = c so by the fundamental uniqueness theorem, fo = c. If Z is a compact subinterval of (a,/?), let K(Z) = {refe:1 I r I2, 8 E I } , a compact subset of S . Given e > 0, there exists then N = N ( l , e) such that (3)

Ifn(z) - cI

<

Vn 2 N,VZ E K(Z).

e

In particular, (4)

If(refe)- CI

<

e

Vr 2 N,VO E Z.

Indeed, given such 6 and r, let n be the integer such that n I r < n + 1. Then the number z = (r/n)efebelongs to K(Z) and n 2 N. Since f(refe)= fn(z), the inequality in (4) follows from (3) and the proof is finished. Exercise 12.31 (LINDELUF[1915]) Let 0 < a In, S = {refe:r > 0, -a < 8 < a}, s" = {reie:r > 0, - a 5 8 < a} and suppose that f: s-t C is bounded, continuous, holomorphic in S and satisfies (I)

lim f(f(re-la)= 0.

r+m

Show that

(2)

lim f(refe)= 0 uniformly for 8 in any compact subset of [-a, a).

r-m

Hints: Introduce So = {refe:r > 0, - 4 2 < 8 < 4 2 )

and form

8 5. Sectorial Limit Theorems

443

This function is continuous and bounded in so, holomorphic in So. Since F(re-'"Ia) equals f(re-la) times a bounded factor and F(reta@)equals times a bounded factor, we see from (1) that lirn F(re*{"Ia)= 0.

r- m

Therefore by 5.16, limr+ F(refe)= 0 for each 0 E [ -a/2, a/2]. In particular,

0 = lirn F(r) = limIf(re-ia'a)la r-r a

I + co

and so

(3)

lim f(re-'"Ia) = 0.

r-00

From ( l ) , (3) and 5.16 it follows that

(4)

lim f(reto)= 0 uniformly for 0 E [-a, - a / 2 ] ,

r- m

while from (3) and 12.30 it follows that (5)

lirn f(refe)= 0 uniformly for 0 in any compact subset of

r- m

(-a, a).

From (4) and (5) the claim (2) follows.

Exercise' 12.32 Give an ab initio proof of 12.31 using the harmonic majorant technique. Hints: It is convenient to put the problem in the following form: Let H =

{ z E @: Im z > O } , f : H u (0,m) --+ @ bounded, continuous and holomorphic in

H. Suppose that f ( x ) -+ 0 as real x -+ +oo. We are to show that

(*)

lirn f(z)

I 4 -.m

=

0

+EHg

for each set He = {refe:r > 0,0 I 0 < (1 - ~)-x}, e > 0. We may assume that If1 I1. Let e > 0 and 0 < S c I be given. Choose xd > 0 so large that (1)

If(x)l I 6 vx 2

Xd.

Let L denote the holomorphic logarithm in @\(iy:y I0) which satisfies

(2)

L(refe)= log r

+ iB

--x 311 for r > 0, - < 0 < -. 2 2

(Cf. 3.43.) Look at the function (3)

1 h(z) = 1 - ; Im L(z - xb), z E H\(xd}.

Omitted Values and Normal Families

444

It is harmonic and bounded in H, equals 0 on (-coy xd)and equals 1 on (x,, co). Consequently the function log(max{)f

1,

6))

- (log S)h = log 8 + log+l - (log S)h

is bounded and subharmonic in H (by 5.26(ii)), and has non-positive limit superior at each point of aH\{x,} = (--00, xd)U ( x d y a).Consequently, by 7.15 this function is non-positive throughout H : (4)

log(max{ If 1,’8}) I (log 8)h on H.

Since Im L(rete - x,) = Im L(eio - x,/r) [for all r > xd, 0 I B I T]and since there exists [by uniform continuity of L on the compact set R n C(0, l)] an X ; > xd such that

+ e.rr

Im L(eie - x d / r ) I Im L(efe) 2 = B

+m

Vr 2 x;,O I B i

w,

it follows that

(5)

h(z) 2

; Vz

E H,

with lzl 2 xi.

From (4) and (5), remembering that log S < 0, we get log(max{If ( z ) l , 8)) I (log 8 ) 2 Vz E H , with IzI 2 xi. Exponentiating gives us finally

If(z)l

I Su12 Vz E H,with Izl 2 xi.

Since limht0SE12 = 0 for each fixed e > 0, the claim (*) is established. Exercise 12.33 Deduce both 12.30 and 5.16 from 12.31. Exercise 12.34 Try to deduce 5.56 from 9.20 and 12.30 by a reflection argument like that used in the proof of Milloux’s theorem 5.50. (See p. 26 of CARTWRIGHT [ 19351.)

If we restrict to “non-tangential” approach, then the method of 12.30 can be further exploited to yield (cf. 5.29(ii))

8 5. Sectorial Limit Theorems

445

Exercise 12.35 Let a < j?, S = {reie:r > 0, a < 0 < j3}, f a holomorphic function in S for which some pair of complex numbers is absent from f ( S ) . Suppose that there exist a < a' < /3' < such that (*)

sup inf{lf(peie)I : a' < 6 < 8')--f 0 as r --f m. o>r

Show that limr+mf(reie) = 0 uniformly for 0 in any compact subset of (a,j?).

Hints: (MONTEL [1917], p. 20) Define fn(z) = f(2"z), z E S,n = I , 2, - .. and K = {reie:1 Ir I2, a' I6 5 p}. For each r E [l, 21 select 6,(r) E [a',j?'] such that

(**I

lfn(reieJr))l= I f(2nreieJr))l= inf{If(2"reiB)I: a' < B < j?'} +=

0 as n --f co, by (*).

This shows that no subsequence of {I fnI} can diverge to 00 un$ormIy in the compact set K. However, by 12.25 { f n }is a normal family in S. We infer that {f,} is locally uniformly bounded in S (see 12.24). Using the Cauchy integral formula in small disks, we further infer that { f n } is locally uniformly equicontinuous in S. In particular, { f n }is uniformly equicontinuous on K and so from (**) it follows easily that if g is the limit of a convergent subsequence u;l,} of cfn},then g(z) = 0 for each cluster point z of {rei8@:n = nl, n,, . . .}. We have this for each r E [l, 21, so such z constitute an infinite subset of K and by 5.62, g is the 0 function. By 7.8 {f n }therefore converges to 0 locally uniformly in S. The proof is concluded from this point just as was 12.30. Exercise 12.36 Use the method of 12.30 to prove this result of LINDEL~P [1909]: i f f is holomorphic in an open sector S, never 0 nor 1, and is bounded on some half-line in S, then f is bounded in every closed subsector of S. Next we are going to investigate the analog of 12.30 when I f(reie)>lconverges as r -+00, rather than f(reie) itself. (An easy result along these lines is 7.31.) Also we will look at a harmonic function analog of 12.30. Here it is more convenient to work with strips and half-strips rather than sectors, so we begin with Exercise 12.37 (i) Check that 12.30 remains valid ifthe sector S there is replaced with a truncated sector S\n(O,R ) for any R > 0. [No modijications whatsoever are necessary in the proof fi one first re-scales to get R < I .] If a, b, c E R with a < b, then there is an exponential which maps the (ii) semi-strip (a, b) x (c, m) conformally onto a truncated sector S\D(O, R). Using it, formulate a semi-strip version of 12.30. Exercise 12.38 Let a, p, a, 6 , c E R with a < a < b < /3 and form S Is> x (c, co). Let f : S -+62 be holomorphic and bounded, by K, say. (i) Suppose that (a,

(I)

lim 1f ( a

y-rm

+ iy)l =

limIf(b

y-rm

+ iy)l = 1,

=

Omitted Values and Normal Families

446

- a < # - a),

(2)

b

(3)

a - a ~ b - a and 8 - b ~ b - a .

Show that for any 6 > 0 there exists c(6) > c such that (4)

1 + iy)l 2 2K

If(x

wheneoer a

+ 6 Ix I

- 6, y 2 c(6).

Hints: For z E S, = (a, 2a - a) x (c, 00) define g(z) = f(2a - z). Note that 2a - z = 2a - 2 E S, 5 S, so g is well-defined and holomorphic. Then F = fg is holomorphic and bounded in S, and satisfies F(a

+ iy) = If(a + iy)l'+

1 as y

--f

a,by (1).

Since a < a < 2a - a, we may cite 12.37(ii) to conclude that limu+mF(x + iy) = 1 uniformly for x in any compact subset of (a, 2u - a). In particular, there exists c1(6) > c such that (5)

IF(x

+ iy)l

>

f V(x,y ) E (a + 6, 2a - a - 6)

x (cl(6), co).

Because If1 I K, (5) and the definition of F show that (6)

If(X

+ iy)l > uy1

V(X,y ) E (a

+ 6,2a - a - 6)

x (cl(6), 00).

Argue similarly to produce c2@) > c such that (7)

If(x

+ iY)l

>

1

V(XY Y ) E (2b - B

+ 6, B - 6)

x

(c2(%

00).

Now by (2), if 6 is sufficiently small, the strips (6) and (7) overlap and so for c(6) = max{cl(6), c,(S)} the inequalities (6) and (7) yield (4). Under hypotheses (I), (2), (3) show that (ii) (8)

lim If(x

u-rm

Hints: Let (9)

If@

e

+ iy)l

= 1

uniformlyfor x in any compact subset of(a, B).

> 0 be given. There is then by (1) a y(e) > c such that

+ iy)l I 1 + e , If ( b + iy)l

I1

+e

Vy

L y(e).

The function

is bounded by I and on [a, b] x {y(e)} it is bounded by 1/K.Consequently, because of (9), the holomorphic functionfh, is bounded in this set by K and bounded by I e on its boundary. From 5.13 it follows then that,fh, is bounded by 1 + e throughout this set. Since limv+mh,(x iy) = 1 uniformly for

+

+

0 5. Sectorial Limit Theorems

447

+

x E [a,b], we may select yl(e) > y(e) sufficiently large that Ih,(x iy)l > (1 e)/(l 2e) for all y 2 yl(e) and all x E [a, 61. We shall then have

+

+ If(x + ir)l I1 + 2~

W, v) E [a,bl

x [yl(.),

00).

Because of (4), the same argument may be applied to llf. Conclude that

(10)

lirn If(x

u-m

+ iy)] = 1

uniformly for x E [a,61.

Now consider 6 > 0. In (i) we saw that limu+mF(x x E [a + 6, a]. Thus, by definition of F, (11)

lim If(x

u-m

+ iy)l lf(2a - x + iy)l

+ iy) = 1 uniformly for

= 1 uniformly for x E [a

+ 6, a].

+

(3)

Since a I 2a - x I 2a - a Ib for x E [a 6, a], it follows that for these x the behavior of the second factor on the left side of (1 1) is governed by (1 0). It follows therefore from (1 1) that lim [ f ( x + iy)l

U-m

=

1 uniformly for x E [a

+ 6, a],

and similarly for x E [b, P - 61. As 6 > 0 is arbitrary, (8) is established. (iii) Prove that (8) followsfiorn (1) and (2) alone.

Hints: Choose a. E [a, a), Po E (6, /I]so that all the hypotheses of (ii) hold in So = (ao,Po) x (c, 00). Set do = - ao), a, = max{a, a. - do}, jl, = mi@, Po + do}.Then if points a,, b, with a, < b, are appropriately selected in (ao,Po), part (ii) will be applicable to S, = (a,,8,) x (c, 00). Construct successively strips So c S1c . - c S in which (ii) applies and observe that S, = S must occur after some finite number n of steps.

+w0

Exercise 12.39 Let a, P, a, b, c E R with a < a < b < jl, F: (a,P) x (c, 00) a bounded holomorphic function. Suppose that (1) (2)

lirn IF(a

u-*

+ iy)l = A

> 0,

lirn IF(b

u-m

+ iy)l

=

--f

C

B > 0,

b - a < +(,9 - a).

Show that lim,,,lF(x + iy)l exists for each x E (a, P), uniformly in compact subsets of such X. Find the value of this limit.

Hint: Show that the function f ( z ) = A-(b-Z)l(b-a)B-(z-a)l(b-a)F(z) fulfills the hypotheses of 12.38(iii). Exercise 12.40 Let a , P, c E Iw with a < 8, F: (a,P) x (c, 00) + C a bounded holomorphic function. Suppose that there are three distinct numbers xl, x2, x3 E (a,P) such that limv+m[F(xk iy)l exists for k = 1, 2, 3. Prove that then lim,,,IF(x + iy)l exists for every x E (a, B) and has the form eax+*for appropriate real constants a and b.

+

Omitted Values and Normal Families

Ma

Exercise 12.41 (i) Suppose a < a < b < t9 and h : S = (a,p) x (c, a)--f 88 a bounded harmonic function satisfying lim h(a + iy) = A, lim h(b + iy) = B. u-

+

u-00

Show that limu+mh(x + iy) = B(x - a)/(b - a) A(b - x)/(b - a)for each x E (a,P). Hints: Replace h(z) by h(z) - B(Re z - a)/@ - a) - A(b - Re z)/(b - a) to achieve A = B = 0. Select (10.2) F E H ( S ) with h = Re F and form f = eF. Since both f and 1/f are then bounded, the first part of the proof of 12.38(ii) shows, without appeal to 12.38(i), that limu-mlf ( x + iy)l = 1 for all x E [a,b]. Now cite 12.40. For an alternative short and elegant Beweisschluss at this point [1926]. see HARDY (ii) (LOOMIS[1943]) Let h be a positive harmonic function in the open right half-plane. Suppose that - 4 2 < 8, < 8, < 4 2 and limtroh(retel)= A , limr,oh(reiep) = B. Show that limrroh(rete) = B(O - 0,)/(8, - 8,) + 419, - 8)/(8, - 8,) for every 8 E ( - ~ / 2 , ~ / 2and ) uniformly for each compact set of such 8. Hints: According to 5.29, h is bounded on each set S, = {rete:0 < r < 1, 181 < n/2 - e} (e > 0). The exponential function E(z) = elz maps (-7r/2, n/2) x (0, co) conformally onto So and each vertical half-line (8) x (0,co) onto the segment {refe:0 < r < I}. Therefore h E fulfills the hypotheses of (i). 0

Exercise 12.42 Let f be entire and suppose that for each real 8 we have either lim If(rete) - c(8)l = 0 for some c(8) E C

(*I

{

or lim lf(rete)l = 00. r-a

(Examples of such functions will be produced in Chapter XV.) Show that in any non-degenerate interval I there is a non-degenerate sub(i) interval J such that If (rete)I 1 1 for all 8 E J and all suficiently large r either (1 .J) I f(reie)l I 2 for all 8 E J and all suflciently large r. or (2.5) Hints: Suppose that no such J exists. Construct non-degenerate closed intervals J, c I and r, > n inductively as follows: given J2n-1,the failure of (I.J2n-1) means O , , E J ~ , - ~and ran > 2n exist such that If(r2,eie1n)I < 1. B y continuity there is then a whole (non-degenerate) subinterval Jan of .I2,,-, such that (1.n) If(r2,ete)I < 1 V 8 E Jan. 1 Similarly, given J,,,, the failure of (2.J2,) means ,8, + E Jznand r2, + > 2n exist such that If(r2,+le*ean+l)( > 2. By continuity there is then a whole (nondegenerate) subinterval Jan+ of Jan such that

+

(2.4

If(r2,+lete)I > 2

veEJ2n+l-

8 5. Sectorial Limit Theorems

449

n&l

For any 0 E J,, (and there is at least one, by compactness) (1.n) and (2.n) each hold for all n. It follows that limn+m~f(r,,e~e)~ does not exist in [O,oo]. Since r, --t 00, this contradicts (*). Call a non-degenerate interval I a convergence sector u o r f)if (ii) either lim sup[f(pefe) r-m

P

~

I

eel

CI = 0

for some c E C

Show that every non-degenerate interval contains a convergence sector.

Hint: Because of (i) and (*), 12.37(i) is applicable to one o f f o r Ilf. Exercise 12.43 Let a < 6, c < d be real numbers, R = (a, b) x (c, d ) and f continuous on R, holomorphic in R. Prove that if min [Ref(b

+ iy) - Ref(a + iy)] 2 b - a,

max [Imf(x

+ id) - Imf(x + ic)] 2 d - c.

csysd

then aixlb

Hints: For each sufficiently small e > 0 form R, = (a

+

E,

b - E ) x (c

laRe

+

d-

E,

a).

By Cauchy’s theorem f = 0. Let E J. 0 and appeal to the (uniform) continuity o f f on R to conclude that f = 0. Writing out this integral and equating its imaginary part to 0 yields.

laR

[ [Ref(b + iy) - Ref(a + iy)]dy d

whence (d - c ) min [Ref(b csysd

=

1’ a

[Imf(x

+ id) - Imf(x + ic)]dx,

+ iy) - Ref(a + iy)] 5 (b - a) max [Imf(x abxib

+ id) - Imf(x + ic)].

Exercise 12.44 Let f be holomorphic in S = (a, b) x (0, a),continuous on 3; and suppose that Kudm Re f(a + iy) < limu+mRef(b + iy). Then f is unbounded in S. Hints: ( P ~ L Y[1933]) A By translating and scaling we may suppose that

lim

u-m

Ref(a

+ iy) < a < b -=

lim Ref(b u-m

There exists then c > 0 such that (*)

Ref(a

+ iy) c a

and Ref(b

+ iy) > b

+ iy).

Vy 2 c.

Omitted Values and Normal Families

450

Set m = max,,,,,lf(x By (*) we have min [ R e f ( b

crurd

+ ic)[and let M > 0 be given. Consider d = c + m + M .

+ iy) - Ref(a + iy)] 2 b - a

and therefore from the last exercise max [Imf(x

arxrb

+ id) - Imf(x + ic)] 2 d - c.

It follows then from the definition of m that max If(x

arxrb

96

+ id)[ 2 d - c - m = M.

Applications to Iteration Theory

Exercise 12.45 Prove that in 7.37-7.41 the hypothesis that the region R be bounded can be weakened to the simple assumption that C\Rcontain at least two points.

Hints: Examine the proofs of these results to see that boundedness of Q is

only used to ensure that the iterates of the self-mapfconstitute a normal family. Since our assumption is that there are at least two distinct points a, b in @\R, we have a, b $frnl(R) c fl for all n, so the desired normality is a consequence of 12.25. One subtle point occurs in the proof of 7.37. We use normality to choose the mjk so that {f["k'} either diverges to 03 locally uniformly on R or else converges locally uniformly on R to a holomorphic limit 0.We have to show that it is the second alternative which occurs. Since is assumed nonconstant, +(a)c R. Indeed, if +(zo) = wo $ R for some zo, thenf["d - wo are never 0 in R but converge locally uniformly on R to - w o which is 0 at the point zo. By 7.1 I it would follow that is constant. Now we may fix a z E R and then {+(z)}u {fW(z):j = 1,2,. . .} is a compact subset K of R. The identity f["~klcfc"fkl(z))= f C 4 k+ l'(Z)

+

+

+

then shows that (f["jk'} does not diverge uniformly to 03 on K. Hence the first alternative for (f'"lk1} does not occur and the proof proceeds as in Chapter VII. To get the needed local boundedness of the sequence of iterates in 7.40, cite 12.24.

Here is the ultimate generalization of the above:

Exercise 12.46 Let R be a region such that C\fl contains at least two points. Let f E H(R) a n d f ( Q ) c R. Suppose that for some zo E R the sequence {f["'(zO)} has a cluster point wo in R. Show that then either f is one-to-one and onto or else f(wo) = wo and (frn1} converges locally uniformly on R to wo.

7. Ostrowski's Proof of Schottky's Theorem

45 1

Hints: If some iteratef["] has a fixed point a in R, then 12.45 may be applied to the functionf["l to conclude that either If[""(a)I < 1 or elsef1"1 is one-toone and onto. The latter clearly implies thatfitself is one-to-one and onto. The former implies by 7.35 and normality (see the Hints to 12.45) that the iterates off[*] converge to a throughout a. In particular, f(a)

=

lim f(f"("l(a)) = lim f[k"l(f(a)) k- m

k- w

= a.

Then by a simple computation with the Chain Rule,f["]'(a) = [~'(cz)]",so the assumption If["l'(a)l < 1 means that If'(a)l < 1. Another appeal to 7.35 and normality leads to the conclusion that {f["]} converges in R to a. But then a = limf["l(z,) must coincide with the cluster point wo. Therefore the proof is reduced to the case where

(I)

no iterate offhas a fixed point in R.

Now mimic the proof of 7.37. Choose n, < n2 < . - so that f["il(zo) -+ wo and set m, = n,+, - n,. Since {f["'} is a normal family, {mr}has a subsequence {m,,} such that either {f["fkl} diverges to 00 uniformly on compact subsets of Q or converges to a holomorphic function 0.Argue as in the hints to 12.45 that the first alternative cannot obtain : Since

(2)

f["~kl(f["~kl(Zo))

= f["ik

+

"(ZO),

the sequence {f["jkl} cannot diverge uniformly to 00 on the compact subset K = {wo}u {f~"i](z,):j= 1, 2 , . . .}. Next, go to the limit in (2) uniformly on K to see that

(3)

W w o ) = wo.

The assumption ( I ) means that f["l - I has no zero in CI for any n, where 1 denotes the identity function. Since f["jkl - 1 converges locally uniformly on R to the function 0 - I which does have a zero by (3), we learn from 7.1 I that 0 - I = 0. It follows then from 7.36 thatfis one-to-one and onto.

97

Ostrowski's Proof of Schottky's Theorem

Exercise 12.47 Let f be holomorphic in D(0, r ) , f ( O ) = I and 0 < Then for all 0 < p < r

If(z)l 2

M-2"'('-")

If1

5 M.

vz E D(0, p).

Hint: Let F be a holomorphic logarithm for Mlfand apply the upper Harnack inequality (5.28).

Omitted Values and Normal Families

452

Lemma 12.48 Let f be holomorphic in a neighborhood of a(0, I), f.(f - 1) zero-jree and for all 0 < r < 1 define M ( r ) = rnaxl f(z)l, IzI sr

1

m(r) = rnax -

c = max{l, Iloglf(0)l

lzlsr

IfWl

I).

Then for any 0 < p < r < 1 (*)

fi(p) 5

elOc/(r-P)[logf i ( r ) ] 2 ’ ( r - , ) ,

Proof: Fix such an r and p. Notice that &(r) 2 e and c 2 1, so IOc/(r - p) 2 10/(r - p ) > 10. Consequently the right side of (*) exceeds e and the proposed inequality is true if f i ( p ) Ie, that is, if m(p) 5 e. So we need only address ourselves to the case where

Let n be the positive integer specified by (3)

n Ilog B(r)< n

+ 1.

Now f is holomorphic and I - f is zero-free in V = D(0, R) for some R > 1. Therefore by 5.35 there is a holomorphic function F such that (4)

Fn = 1 - f in V .

We set f * = n( 1 - F ) . Let s denote the holornorphic nth root in D( I , 1) which is positive on (0, 2), that is, s = eLinwhere L is the Principal Branch of the Logarithm in C\( -m, 01 (3.43). We will show that F can be chosen so that (5)

F

= s 0 (1

- f)

near z,.

Indeed we have (6)

If(z,)l =

1 1 1 I- < m(p) e 2

so for some connected neighborhood U of z , in V we have If(z)l < 1 Vz E

u.

$ 7 . Ostrowski's Proof of Schottky's Theorem

453

For these z then F(z) and s(1 - f(z)) are each (non-zero) nth roots of 1 - f(z) and so their quotient is an nth root of 1. As s 0 (1 - f ) / F is continuous on the connected set U,it is therefore constantly equal to some one nth root of 1, say ezni"". We may replace F by eZni'lnFwithout prejudice to (4) and upon so doing we shall have (5) valid throughout U.Computing dk)as [s I(0, 2)](lC)and remembering that on (0,2), s is just the unique positive nth root function, we see s(k)(l) = s(w) =

1n (An - 1 ) . . . (i - k + 1)

" 1 12 (-

1

k=On

-k

+1

-

1)k VWED(1,l).

It follows that ldk)(l)1 I (k - I)!/n and so

2 < - Iw - 1 ) V w E D ( I , + ) * n

(7)

Thus for z near z, we have from (3,(6) and (7) If*(z)l < 2lf(z)l. In particular, If*(z,)l < 21f(z,)l

=

2 q q

by (2)

and so 2 (8)

1

def.

T)

< m*(p).

Moreover, from the definitions of F and f *

+ (I + ~ 2 ( r ) ) l / ~ ] < [log fi(r)][l + (1 + en+l)l/"] by definition (3) of n = [I + e(eWn + e)l/"] log A ( r ) < [1 + e(e-" + e)] log A ( r ) I (e2 + 2) log A ( r )

~ * ( rI ) n[l

(9)

< 10 log A(r).

454

Omitted Values and Normal Families

In the last step we used the fact that e2 c 8 (exercise). Next note that I F(0)l s (1 + I f(0)l)l/nby (4), so by (4) and the definition off *

= I

+-If (10)I I1 + ec,

since c 2 -log)f(O)l by definition,

< enc.

(10)

Now 12.47 applied tof*lf*(O) [note thatf* is zero-free] says that

Using this and (8), (9), (10)

as claimed.

Lemma 12.49 Let the notation and hypotheses be those of the previous lemma. Then

Prooi: First consider only 0E(O, I/e] and suppose, with a view toward ultimately reaching a contradiction, that for some 0 E (0, I/e] (1)

M(I - 8)

,esec(i/e)ioe(i/e).

Now apply the last lemma to l r i n the role off. The conclusion is then (2)

M(p) IelOc'(r-P)[IogG(r)]2/(r-D).

2 (;)'elc

(since c 2 1 and 8 5 l/e)

- 1)

1 > -(elc ea

el - 1 _.- 7c- 1 > 7c _.ea 7

= 7c elc 82

(since (ex - l)/x is an increasing function of x E [l, 00)) a e l - 1 > (;)ac.loo

(from es > 200, above, follows el > 401) 2

> 100cjlog-2 e (since x > logx).

Thus

f i ( l - 8/21

,

e100~(21e)10g(21e)~

The right side exceeds e so the definition of 61 shows that f i ( 1 - 8/2) = m(1 - 8/2) and so our inequality says that m(l - 8/21 > eioo~(a/o)iog(a/e) > e66C(2/8)lOg(2/8)~ (3) Now just as (3) was deduced from (2) via 12.48 applied to 1 If, we can deduce from (3) via 12.48 applied to f that ~ ( -1 8/4) > e66c(4/8)los(4/8)~ (4) The whole implication " ( I ) =- (4)" can now simply be iterated n times (for any positive integer n) and there results the inequality

M(1 - 8/49 > e66c(4"/8)10g(4"~8), n = I , 2, . . . (5) Now (4n/8)log(4n/8)-+ 00 as n +-00 (see 3.18) and so for large values of n, ( 5 ) contradicts the boundedness off in D(0, I). (Remember, f is holomorphic in a neighborhood of b(0, I).) Thus ( I ) is untenable and we have proved that M(j - 8) 5 e66C(l/O)lOP(1/8) , 0 c 8 5 I/e. (6) Now (l/8) log(e/O) is clearly a decreasing function of 8 E (0, I], so its smallest value on this interval occurs at 8 = 1 and is 1. It follows that for 8 E [I/e, I )

Omitted Values and Normal Families

456

Since the right hand of (6) is not greater than the right hand of (*), the conjunction of (6) and (7) yields (*).

Corollary 12.50 (Schottky) Let f be holomorphic in D(0, 1) and omit 0 and 1 from its range. Set d = max(1, loglf(0)l). Then Proof: For each r E (0, 1) the function J(z) = f(rz) is holomorphic in the neighborhood D(0, l / r ) of D(0, 1) and so the last lemma applies to it and &ms that If(rz)l 5 e18ac(11e)10~(ele) Vz E D(0, 1 - 0).

This holding for all r < 1, we let r .f 1 and conclude that (**)

If(z)l

I

e188c(1/e)10g(e18) Vz E D(0, I

- 0).

Here c = max(1, Iloglf(0)l I} and of course d I c. If d = c, we have in (**) (a bit more than) the desired inequality (*). If d < c, then d = 1 and c > 1. Thus loglf(0)l # Iloglf(0)l I, so loglf(0)l = -Iloglf(O)I I = - c < -1, If(0)l < l/e < Then certainly e > 11 -f(O)l > l/e. Consequently the appropriate c-constant for the function I - f is 1 and (**) for this function reads

+.

11

- f(z)l

I

e18a(11e)10~(e/e) Vz E D(0, 1 - 0),

whence

~f(~)l

I 1 + eisa(iie)~os(e/e)<

vz

e18B(l/~)lo~(e/~)

~ ( 01, - 0)

(since ( I / @ log(el0) > I , as noted in the last proof). This confirms (*) in the complementary case d < c and finishes the proof.

Remarks: This is perhaps the simplest of the elementary derivations of Schottky’s theorem which nevertheless give quantitative estimates for the [1925a]. He also majorant, free of unspecified constants. It is from OSTROWSKI presents the same proof in the appendix of his monograph [1931] but with minor modifications which lead to the better inequality

Notes to Chapter XII The trek to the Huuptsutz 12.14 (MIRANDA[1935]) is laborious but, I think, worth the trouble. For one thing, these techniques have a quite broad applicability; they represent the elementary phase of a very fruitful and deep chapter of function theory inaugurated in this century by R. NEVANLINNA. See his book [1953], [I9701 and HAYMAN[1964]. My account follows VALIRON[1937]. I have,

Notes to Chapter XII

457

however, tried to watch over the constants more carefully and to keep before the reader at all times their independence from the particular functions being analyzed. The inequality we obtain is not the best. It can be improved, as regards its dependence on 1/(1 - r), for example, by using techniques from MILLOUX[1940]. HIONG[1958a] proves that for each non-negative integer n there are constants H,,, Kn such that

for allfas in 12.14. In its dependence on 1/(1 - r ) this is a marked improvement over the inequality in the text. Contrary to that author’s claim (p. 987, op. cit.), I found that when the proof details and supporting material had all been supplied, the proof of (*) was longer than that in the text. However, a more serious disadvantage of (*) is the fact that the bound onfwhich it provides gets coarser as If(0)l < 1 gets smaller, a circumstance somewhat at variance with intuition, and because of this I was unable, despite the author’s claim (p. 994, op. cit.), to deduce from (*) the important normality criterion 12.25. [For the history and priorities on the latter see p. 987, op. cit.] So I settled for the version of Miranda’s Theorem in the text. The reader may wish to investigate this matter on his own. He will find in HIONG[1968] an English translation of (a Chinese translation of) HIONG[1958a] and interesting related matter in HIONGand Ho [1961], SHIEH[I9621 and YANGand CHANG[1965]. [1904], The special case of the theorem in 12.15 and 12.50 is due t o SCHOTTKY [1905], [I9061 independently, though his deriva[1906], [1917] (and BOUTROUX tion contains a non-trivial lacuna), and is already a very deep and striking result. His original theorem asserted a bound on log M(r,f) dependent only on f(0) and (1 - r)-4. LANDAU[I9061 improved this to (1 - r ) - 2 and finally LBVY[1912] showed that the dependence on r was of the order (1 - r)-l. For Schottky’s theorem there exist very polished direct proofs, the products of many hands: LANDAU[1926], BLOCH[1924], [1925], VALIRON[1926], [1930], H . KBNIG [1957], ESTERMANN [1971]. See, for example, the account in SAKSand ZYCMUND[1971], pp. 341-350(cf. MACKI[1968]) or CONWAY[1973] or LANDAU [I9401 or [1929a]. (See especially the latter, pp. 18-19 for history.) For a some[1939], CARTWRIGHT [1956] and HOLLAND what different proof see TITCHMARSH [1973]. For an exhaustive account of Schottky’s theorem exploring many of its [1931], chapters 1 and 4. For exapplications see the treatise of OSTROWSKI tensions of a different kind from Miranda’s see BIEBERBACH [1922], LANDAU [1922b], FEKETE [1922], SAXER[I9341 and KRAJKIEWICZ [1977]. In our version neither the constants nor the order of dependence on I /( I - r ) are best possible. In HAYMAN[I9471 occurs one of the sharpest forms of the inequality :

458

Omitted Values and Normal Families

JENKINS [1955] gives a short proof of this and some improvements which de[1938] had proved a corresponding pend on the size of If(0)l.Earlier AHLFORS inequality with the (larger) constant 4 + log 10 instead of 7r. See also ROBINSON [1939]. In Chapter XVIII we will prove a sharp Schottky-type inequality for univalent functions. OSTROWSKI [ 19331 finds asymptotic expressions for M(r, f) as r.T 1. The amazing lemma 12.10 is from BUREAU[1932]; the ideas in it go back to BOREL[1896], [1896-971 and BLUMENTHAL [1910]. A version of it features also in Ostrowski's proof. (See the proof of 12.49.) 12.16 is due to BUREAU[1931] and 12.17 to PICARD[1879a]. The beautiful corollary 12.18is asserted inP6LYa [1921]and [1922]and proved inSAXER,[1923], VAROPOULOS [1928] and CSILLAG [1928], [1935]. This result has a remarkable generalization due to TUMURA [1937] (whose proof contained gaps) and CLUNIE [ 19621: iff is entire and for some k 2 2, f . f ( k )is zero-free, then f has the form [1964]. For a f(z) = eaz+bfor some constants a and b. See pp. 67 ff. of HAYMAN short, but not elementary, proof when k = 2, (first proved by Hayman) see C.-c. YANG[1970]. See also S. S. MACINTYRE [1949] and EDREI[1955]. Finally, FRANK [1976] has proved a conjecture of Hayman that if meromorphic f are admitted, the only new members of the club aref(z) = (az + b)-",n a positive integer. For 12.19 see ROSENBLOOM [1948] and [1952]. It is pointed out in the review of the first paper that (part of) the result is actually contained in an earlier one of FATOU [1926] (p. 346). For an interesting application of 12.19 see the American Mathematical Monthly 73 (1966), 404-405 and for a modest extension of 12.18, the same journal, 66 (1959), 155. 12.20 is from chapter IV of MONTEL [1916]. (See the last paragraph of the review of this paper.) A little more is true: if p, q are integers and l/p + l/q < 1, then there are no non-constant entire functions F, G such that FP + GQ= 1. See VALIRON [1929a], JATEGAONKAR [ 19651 (where the proof follows the pattern of 12.20) and GROSS[1966]. Actually 12.20 is a special case of a result of Picard affirming that if Pand G are meromorphic in D(0, l)\{O} and satisfy an algebraic relation of genus greater than 1, then 0 is a removable singularity of Fand of G. [One takes P(z) =f(l/z), G(z) = g(l/z).] [1904], with a 2 1/58. LANDAU [1906], p. 285 12.21 is Satz VIII in HURWITZ gave a proof of this also. 12.21 was rediscovered by BOCHNER [I9261 and FEKETE [ 1927al. CARATH~ODORY [ 19071 and Bochner each confirmed Hurwitz' conjecture that 1/16 is the best value for a. (See also NEHARI [1952], p. 323 or SANSONE and GERRETSEN [1969], p. 502.) The proofs use a modular function; see below. For a related result see LANDAU [1922a], BOHR[I9231 and MONTEL [19291. The important special case of 12.25 where N = 0 is due to MONTEL[I9121 (but is adumbrated in Sutz VI of CARATH~ODORY and LANDAU [1911]; see also

Notes to Chapter XI1

459

the footnotes there.) For another proof in that case see DE LA VALLBE POUSSIN [1915-161. This theorem is the principal ingredient in our development of the Julia theorem (12.27); in turn its principal ingredient is Schottky’s theorem. If we let our functions take their values on the Riemann sphere C, (see Chapter I notes), the idea of normal family extends naturally to meromorphic functions and becomes synonymous with the modem topological idea of pre-compact set (in the appropriately topologized space of functions). This was first pointed out by OSTROWSKI [1926b], Q 3. In terms of the so-called chordal metric in C, a very [1964], pp. useful normality criterion was given by F. Marty. (See HAYMAN 158-160 for references and a statement and proof of the result.) This can be made an alternative starting point, from which Montel’s criterion, thence Julia’s theorem and (a qualitative version of) Schottky’s theorem can be deduced. For the elegant details see ZALCMAN [1975]. The important paper of CARATH~ODORY [1929b] should also be mentioned in this context. MANDELBROJT [1929] shows that for a family 9 of zero-free holomorphic functions in a region Q the local uniform boundedness in Q x Q of the family is necessary and sufficient for of auxiliary functions f(z, w) = f ( z ) l f ( w ) ,f E 9, the normality of 9. Cf. 7.28. For an elementary proof of a slightly weakened version of 12.28 see ZALCMAN [I 9781. Before 12.27, PICARD [1879b] had proved that there exists b E C such that C\{b} c f ( D ( a , R)) for all (sufficiently small) R >.0, i.e., in every neighborhood of a, f assumes all complex values with at most one exception. This is the socalled Big (or Great) Picard Theorem [in contrast to the little (or lesser) one in 12.17.1 See FRANKLIN [1925a] for a modest extension and CARATH~ODORY [I 9 I2b] (also OSTROWSKI [ 1929b], BIEBERBACH [ 19291, LANDAU [ 1929~1and FENCHEL [1931]) for a quantitative extension: there exists a constant r E (0, 1) such that eachfwhich is holomorphic in A(0, 1) and omits 0 and 1 from its range satisfies eitherf(A(0, r ) ) c D(0,2) orf(A(0, r ) ) c C\D(O, +). For Julia’s extension of Picard’s theorem see p. 102 of JULIA[1924]. Interesting extensions [ 19311. An extension to harmonic of Julia’s theorem may be found in OSTROWSKI functions appears in MONTEL[ 1932~1. There is an entirely different, rather geometric, development of Schottky’s theorem and many of its corollaries like the Picard theorems, based on the Monodromy Theorem (Chapter X notes) and a modular function. The latter is a holomorphic map of the upper half-plane onto @\{O, I} which is invariant under a certain group of conformal self-maps of the upper half-plane. For a careful treatment of it see pp. 353-356 of RUDIN [I9741 or chapter 23 of ESTERMANN [1962]. This approach was the historically first one, used for example by Picard himself, and it is via a modular function that the sharpest inequalities have been [ I9061 and the works referenced there, CARATH~ODORY obtained. See LANDAU [1960], DINGHAS[196l], @ 7 l , 72; HILLE [1962], Q 14.5; VEECH[1967],

Omitted Values and Normal Families

460

pp. 126-136 and LEHNER [1969]. The proof given by P~RRON [1929] is a more elementary version of the modular function proofs; he manufactures on the spot a substitute for the modular function. For a more detailed exposition of Perron’s work see $ 140 of PRINGSHEIM [1932]. Cf. also DE LA VALLBEPOUSSIN [1915-16]. After Picard proved his results with the modular function a search began for an elementary proof (meaning one which did not use the modular function or any analytic results of comparable depth). In [1896] BORELachieved this for the little theorem (see also the appendix to his book [1921]) and in [1904], building on Borel’s technique, SCHOTTKYdid it for the great theorem. For a nice treatment of their work (including contributions of Carathtodory and Landau) see the congress lecture of LINDEL~F [1910]. The next phase occurred with the arrival of Bloch‘s theorem 7.26(iii) (see Chapter VII notes), already presaged in the work of Landau, Hurwitz and Koebe. From this follows a relatively easy proof of Schottky’s theorem, thence the Picard theorems. This is the approach followed in almost all textbooks which want to avoid the modular function. For a detailed history of the Schottky and Picard theorems up to 1920, see pp. 409417 of BIEBERBACH [1921]; for related history see VALIRON[1932]. In 12.29 (taken from pp. 66-70 of MONTEL[1927]) the reader has a brief introduction to Montel’s theory of quasi-normal families. See his book just cited and VALIRON[1929b] for more on this topic. 12.30 is due to MONTEL [1912]. Nice generalizations in which we only hypothesize that f(t,) c for a certain sequence of points z, E S may be found in EGGLESTON [1951], A. EVANS[1953] and BOWENand MACINTYRE [1954] (cf. 7.31). For a good summary statement of the classical results, containing 5.16 u 5.56 u 12.36 u 12.31, plus references to important work of W. Gross and F. Iversen on the subject, see p. 420 of the encyclopedia article of BIEBERBACH [1921]. For a unified treatment of all these results via harmonic majorants see pp. 301-308 of TSUJI[1959]. For other proofs of 12.31 see $21 of MONTEL [1917], exercise 111.277 of P ~ L Yand A SZEGU[1972] and pp. 12-13 of F. and R. NEVANLINNA [1922]. For 12.36 see p. 518 of MONTEL [1912] and Ou [1957]. --f

12.38-12.40 are from HARDY, INGHAM and P ~ L Y[1928]. A These authors show that equality can be tolerated in hypothesis (2) of 12.38 and 12.39 and provide a counterexample to the conclusion when b - a > - a). But they show how 12.41(i) can be used to eliminate all hypotheses on a, b (beyond a # b, of course) when the function has no zeros. See CARTWRIGHT [I9621 and HAYMAN [1962] for extensions. 12.41(i) is from HARDY[1926]; for a generalization see TSUJI [1931a], BONSALL[1949] and BEARDON[1971] (subharmonic [1930b], SHIMIZU functions); see also pp. 392 ff. of OSTROW~KI [1937].On 12.41(ii) see YANAGIHARA [1974]. [1965] and pp. 4243 of DONOGHUE

Notes to Chapter XII

461

I took 12.42 from ROTH[1938], though it is adumbrated in BOHR[1930]. Compare also LAURITZEN [1950]. Bohr proves an analog of 12.42 for entire functions f which are bounded on every ray: given any non-degenerate closed interval J c [0,27r], there is a non-degenerate closed interval I c J such that Cf(refe):B E I, r 2 0} is bounded. He also shows that if {I,,} are disjoint open intervals in [0,27r] whose union is dense in [0,277], then there exists a nonconstant entire f which is bounded on every ray and uniformly bounded in each set {refe:tl E I,, r 2 O}. Without the uniformity feature 12.42(ii) is essentially contained in BOHR[1927]. Non-constant entire functions bounded on every ray through 0 are not too hard to construct: see D. J. NEWMAN [1976]. As to non-constant entire functions which have limits at infinity along every ray, these too exist; in Chapter XV we will construct a few. [The easiest example I know of is the function f(z)f(iz), where f is the function NEWMAN constructs, op. cit.] Both 12.43 and 12.44 are taken from P ~ L Y [1933]. A See also AHLFORS[1933], MILLOUX [1935], K E ~ K J A R[1934/35], T~ A. J. MACINTYRE [I9391 and JENKINS [1953]. An even briefer proof of a related result is offered in LANDAUand OSSERMAN [1959]. They use the Cauchy-Riemann equations and a double integral instead of Cauchy's theorem. 12.45 may fail if only one point is omitted. Specifically, for R = C\{-2} and = 2(d - I), we havef(0) = 0 and f(Q)= Q. Since, however,f'(O) = 2, the conclusion of 7.40(i) does not hold here.

f(z)

In 12.46 I have followed HEINS[1941b]. Regarding 12.6 and its converse 7.33, OSTROWSKI [1925b] proved the following definitive result: Let Y = {fE H ( D ) : suprcl loglf(reie)IdB < oo}. Here D = D(0, 1) and the integral is in Lebesgue's sense. A necessary and sufficient conz2,. , .} is dition on points z, E D that there exist f E 9 whose zero set is {zl, (1 - lz,l) < 00. For a related result see DENJOY [1929] (cf. also the that Chapter VII notes). Proofs of 12.6 and 12.7(ii) may also be found on pp. 26-28 of F. and R. NEVANLINNA [1922]. They show that the hypothesis (*) of 12.6 is equivalent to F being a quotient of two bounded holomorphic functions in the disk (this follows easily from 7.14(ii)), so the generality of 12.6 over 6.9(iii) is illusory. Ostrowski's result above (resp., 6.9(iii)) together with 7.33 enable one to factor any function of class Y (resp., H " ( D ) ) into the product of a (bounded) zero-free holomorphic function in D,i.e., the exponential of a holomorphic function in D,and a Blaschke product. This representation is extremely useful. See the elementary discussion on pp. 40-45 of MONTEL[1927]. For a [1970]. comprehensive treatment see chapter VII of R. NEVANLINNA

R"

x;=l

If hypothesis (*) in 12.6 is weakened to then the weaker conclusion (I [1971].

z:l"=l Iz,~)~

:j

log+IF(peie)IdBdp < 00, < oo can be inferred; see KABAILA