Fix Points and Factorization of Meromorphic Functions

and applying (1. 8) to the formula

=n

h (z),

+n

log p

1 2

"

loglf(pei'P)ld
0

[P n(t, j) - n(O, j) dt + n(O, f) log p

pe''P

0

(0, j)

n(r, fJ = n(r, t) -

we get, by making use of the relations (1.10),

r" log+ If(pei'P) Id

211"

Noting that

(0, j) - n(O, j) ,

10 10 _ -.!...1 2 " log + If( 1.) IdIP + -

211"

is also meromorphic in

n(r, j) - n(O, f),

S

-.!...

c..

h (z)

1 P

0

t

n

(t, t) - n (0, t) dt t

+ log Ie. I .

(1.11)

This formula which contains (1.8) as a particular case, is called JensenNevanlinna formula. 1.3. CHARACTERISTIC FUNCTION Let f(z) be a meromorphic function in a domain Izl < R (0 < R ~ 00) non identically equal to zero, and consider the formula (1.11). H in this formula we set p = r and define (1.12)

N(r, f)

=

fo" n(t, j) ~ n(O, j) dt + n(O, j) logr ,

(1.13 )

Nevanlinna's Theory of Meromorphic Functions

7

then (1.11) may be written as (1.14) Next we define

T(r, I) = m(r, I) + N(r, I)

(0 <

r

< R) .

(1.15)

Then (1.14) becomes

T(r, I) = T (r,

-7 ) + log Ic.1 .

(1.16)

Nevanlinna calls T(r, I) the characteristic function of the function I(z). In the particular case that I(z) is identically equal to zero, T(r, I) = 0 is also defined, but (1.16) no longer holds. Now we are going to prove a formula for T(r, I) and deduce some of its properties. We start from the formula (1.17) This formula holds for any complex number a. In fact, this is evident, if a = O. IT a =1= 0, we apply (1.6) to the function ip(z) = a - z with p = 1 and get

In order to obtain the desired formula for T(r, I), we assume that I(z) is non-constant and 1(0) is finite. (} being a real number such that eiO =1= 1(0), application of (1.7) and (1.8) to the function I(z) - eiO yields

11211"

211"

0

.

log I/(re''P) - e''0 Idip

(1)

= N r, - I '0 -

e'

-

N(r, I) + log 1/(0) - e''0 I ,

Fix'points and Factorization of Meromorphic Functions

8

where 0 < r < R, Next keeping r fixed and integrating both sides with respect to (), we get

-1 211"

1211" d()-1 1211" log I/(rei'P) 211"

0

= -1 21r

eioldcp

0

1211" N ( r, - -1 ' 0) d() 1-

0

e'

1211" log 1/(0) -

N(r, f) + - 1 211"

0

eiold() . (1.18)

Then interchanging the order of integration of the integral on the left of (1.18) and making use of (1.17), we get

-1 211"

1211" d()-1 1211" log I/(rei'P) 211"

0

= -1

211"

= -1

211"

1 211"

eioldcp

0

1211" dcp-1 1211" log I/(rei'P) 211"

0

eiold()

0

1211" log+ I/(rei'P)ldcp = m(r,f)

,

0

1211" log 1/(0) -

eiold()

= log+ 1/(0)1

.

0

Finally from these relations and (1.18) we get the desired formula

11211"

T(r, f) = 211"

0

N

1)

( r, - I'0 -

e'

d() + log+ 1/(0)1 .

(1.19)

This formula is due to Cartan H. In the proof of (1.19) we have assumed that 1(0) is finite. If the point z = 0 is a pole of I(z), then by (1.7) and (1.11) we have

11211"

211"

log I/(rei'P) - eioldcp

0

=N

1)

( r, - I'0 -

e'

-

N(r, f) + log le.1 ,

where e. is the coefficient in the development (1.9), and we get in the same way the formula

11211"

T(r,f)=211"

0

N

1)

( r ' - I"0 -

e'

d()+logle.l·

(1.20)

From (1.19) or (1.20) we see easily that T{r, f) is a non-decreasing function of r and a convex function of log r. In fact, it is sufficient to show that

Nevanlinna's Theory of Meromorphic Functions

9

this is true for N(r, I). That N(r, I) is a non-decreasing function of r is obvious. To see that it is a convex function of log r, consider three values rj(j = 1,2,3) such that 0 < rl < r2 < r3 < R. We have

and similarly

Consequently

This implies

(1.21) which shows that N(r, f) is a convex function of log r. Now we are going to prove some inequalities for m(r, f), N(r, I) and T(r, I), which are often used. First of all, it is easy to see that the inequalities p

log+ (0102'" Op) ~

L

log+ OJ ,

(1.22)

j=1 P

log+{OI

+ 02 + ... + Op)

~

L

log+ OJ

+ logp

(1.23)

j=1

hold for arbitrary OJ ~ 0 (j = 1,2, ... , pl. In fact, to see (1.22), it is sufficient to note that the left member of (1.22) is 0, if 0102 ... Op < 1 and IS p

IOg(01 0 2 ... Op)

= LlogOj ~ j=1

P

Llog+ OJ, j=1

10

Fix-points and Factorization of Meromorphic Functions

if ala2 ... a v ~ 1. On the other hand, to see (1.23), let a = max(al' a2, ... ,av ) and note that the left member of (1.23) does not exceed log+ pa ::; log+ a

+ log p .

Let li(z) U = 1,2, ... ,p) be p meromorphic functions in a domain R (0 < R ::; 00). From (1.22) and (1.23), we deduce immediately the inequalities

Izi <

v

m(r, 1112··· Iv) ::;

L m(r, Ii) ,

(1.24)

i=1 v

m(r, 11

+ 12 + ... + Iv) ::; L

+ logp

m(r, Ii)

(1.25)

i=1 which hold for 0 < r < R. On the other hand, if Ii (0) then we also have the inequalities

i 00 (j =

1,2, ... ,p),

v

N(r, II 12 ... Iv) ::;

L

N(r, Ii) ,

(1.26)

i=1 v

N(r, II

+ 12 + ... + Iv) ::; L N(r, Ii) ,

(1.27)

i=1

o< r <

R. We give a proof of these inequalities only for the case p = 2, because then they are true in general by induction. Evidently it is sufficient to show that n(t, II 12) n(t,

::; n(t, Id + n(t, 12) ,

II + 12) ::; n(t, II) + n(t, 12) ,

(1.28) (1.29)

for 0 < t < R. To see (1.28), consider a disk Izl ::; t and first assume that lI(z)i2(z) has no pole in this disk, then (1.28) is evident. Next assume that lI(z)i2(z) has poles in this disk and let Zk (k = 1,2, ... ,q) be all

Nevanlinna's Theory of Meromorphic Functions

11

the distinct poles in this disk of Idz) and 12(Z). For each k, define Sll) as follows: SkI) is equal to the order of Zk, if Zk is a pole of Idz)j otherwise Sll) = O. Similarly define Sk 2) and Sl12) with respect to h(z) and h(z)/2(Z). We see easily that

and hence

This proves (1.28), because the three sums are respectively equal to the three terms in (1.28). (1.29) is proved in a similar manner. It can also be shown that, if R > 1, then (1.26) and (1.27) hold for 1 ~ r < R without the assumption Ij(O) =F 00 (i = 1,2, ... , pl. Finally from (1.24)-(1.27), we get the inequalities v

T(r, hh .. · Iv) ~ LT(r, Ij) ,

(1.30)

j=1

v

T(r,h

+ h + ... + Iv)

~ LT(r,/j)

+ logp

(1.31)

j=1

which hold under the same conditions as for (1.26) and (1.27). For a function I(z) holomorphic in a domain Izl < R (0 < R ~ 00), the functions T(r, J) and

M(r, I) = max I/(z)1 1.. I=r

(1.32)

are both important. It is interesting that they satisfy the inequality

T(r, J)

~ log+

M(r, J)

~

p+r -T(p, J) p-r

for 0 < r < p < R. The first part of (1.33) is obvious, because

N(r, J) = 0,

T(r, J) = m(r, J) .

(1.33)

Fix-points and Factorization of Meromorphic Functions

12

To prove the second part, we make use of (1.1) which now becomes log IJ(z)1

=

~ 211"

+ z) dt.p 10r" log IJ(pei'P) IRe (pe~'P pe''P - z h

-

2-

Ip - a~z I ~ og p(z _ a~) . '" I

In this formula, setting z = Zo where Zo is a point of the circle Izl that IJ(zo)1 = M(r,f) and noting

zo) < p +

+ O < R e ( pei'P . pe''P - Zo

-

r

P- r

=

r such

,

we get

1

zo) dt.p

2" . (pei'P + log M(r, f) ~ -1 log+ IJ(pe''P)IRe. 211" 0 pe''P - Zo p+r ~ -T(p,f). p-r

1.4. FIRST FUNDAMENTAL THEOREM Let J(z) be a non-constant meromorphic function in a domain Izl < R (0 < R ~ 00). Consider a finite value a and the development of J(z) - a in the neighborhood of the point z = 0

J(z) - a

= c.z· + e.+lz·+1 +...

(c. =I 0) .

By (1.16),

T(r, J - a) = T (r, J ~ a) + log le.1

(0 < r < R) .

Let us compare T(r, f) and

T(r, J - a)

=

m(r, J - a) + N(r, J - a) .

We have

N(r, J - a)

=

N(r, f)

( 1.34)

13

Nevanlinna's Theory of Meromorphic i
and

m(r, J - a) :::; m(r, I) + log+ lal + log~~ m(r, I) :::; m(r, J - a) + log+ lal + log 2 . Consequently

Im(r,J - a) - m(r,l)l:::; log+ lal +log2 and

IT(r, J - a) - T(r, 1)1 :::; log+ lal + log 2 .

(1.35)

(1.34) and (1.35) yield

T

(r, J ~ a) = T(r,1) + h(r) ,

(1.36)

where

Ih(r)l:::; ilog Ic.11 + log+ lal + log2 . Nevanlinna calls (1.36) the first fundamental theorem. Furthermore, consider a function of the form

F(z) _ aJ(z) + b - cJ(z) + d ' where a, b, c, d are constants such that ad - be i- o. If we regard F as a fractional linear function of J, it can be decomposed into several functions of the following forms

JI(z) = J(z) + (3,

h(z) = aJ(z),

1

h(z) = J(z) .

where (3 and a i- 0 are constants. Since the characteristic function of each of these functions differs from T(r, I) by a bounded function, we get the following result: The difference T(r, F) - T(r, I) is a bounded function. This is a complement of the first fundamental theorem. Finally we make an investigation of the growth of T(r, I), when J(z) is a merom orphic function in the <;omplex plane. Such a function J(z) is briefly called a meromorphic function, as already mentioned at the beginning of this chapter.

Fix-points and Factorization of Meromorphic Functions

14

We first prove that if J(z) is a non-constant meromorphic function, then

= 00

lim T(r, f)

r--+oo

In fact, if J(O)

= 00,

•

(1.37)

then by the definition (1.13) of N(r, J)' evidently lim N(r, f)

r--+oo

= 00

and a Jortiori (1.37) holds. If a

= J(O) "I

00,

then

lim T

(r, - J

) =

00 •

r--+CX)

1

-

a

So (1.37) again holds by (1.36). Next we are going to prove that if J(z) is a transcendental meromorphic function, then lim T(r, f) = 00 . (1.38) r--+oo log r Distinguish two cases: 10 J(z) has no pole. In this case, J(z) is a transcendental entire function. In its development n=O

there is an infinite number of coefficients different from zero. It follows then from Cauchy inequalities

lanlr n

~

M(r,f)

(r> O,n

= 0,1,2, ... ) ,

that lim M(r, f) r--+oo

= 00

rP

for each positive integer p. This implies lim log M(r, f) = log r

00 .

r--+oo

On the other hand, taking p = 2r in (1.33), we have log+ M(r, f) ~ 3T(2r, f) . Hence (1.38) holds.

(1.39)

15

Nevanlinna's Theory of Meromorphic Functions

20 f(z) has poles. First assume that f(z) has an infinite number of poles. Then from

N(r2, J)

~

N(r2, f) - N(r, f)

~

n(r, J) log r

(r > 1) ,

we have lim N(r, J) = log r

00

r-+oo

and, a fortiori, (1.38). Next assume that f(z) has only a finite number of poles bj(j = 1,2, ... ,k) whose orders are respectively mj(j = 1,2, ... ,k). Set k

P(z) =

II (z -

bj)mj,

g(z) = P(z)f(z) .

j=1 Then, remembering that f(z) is not a rational function, g(z) is a transcendental entire function, and hence (1.38) holds for g(z). On the other hand, by (1.30),

T(r, g)

~

T(r, P) + T(r, J)

~ mlogr

(r ~ 1) ,

+ K + T(r, J)

where m = E~=1 mj and K > 0 is a constant. Hence (1.38) also holds. Concerning the growth of a meromorphic function, an important notion is that of its order. The order p of a non-constant meromorphic function f(z) is defined by p = lim log T(r, J) . (1.40) r-+oo log r We have 0 ~ p ~ 00, and we shall denote it by p(J). When p is finite, then, for each positive number e, on one hand there is a value ro such that

T(r,J)

<'-pH

for

r > ro,

and on the other hand there is a sequence of values r,. (n to 00 such that

=

1,2, ... ) tending

T(r, J) > ,-P-. for r = r,. (n = 1,2, ... ) . By a complement of the first fundamental theorem given above, for a function of the form F(z) = af(z) + b ,

cf(z) + d

Fix-points and Factorization of Meromorphic Functions

16

where a, b, c, d are constants such that ad - be -I- 0, the difference T(r, F) T(r, J) is a bounded function_ Consequently the functions F(z) and J(z) have the same order. Now consider the particular case that J(z) is a non-constant entire function. Then by (1.33),

T(r, J) ::; log+ M(r, J) ::; 3T(2r, J) and therefore lim log T(r, J) log r

=

r--+oo

lim log log M(r, J) . log r

r--+oo

This shows that, in the particular case of entire functions, the present definition of order is compatible with the classical one. For a meromorphic function f(z) which is a constant, we define its order p(f) = 0. 1.5. LOGARITHMIC DERIVATIVE Let us return to Poisson-Jensen formula (1.1) and consider a point Zo of the disk Izl < p such that f(zo) -I- 0,00. Then J(z) is holomorphic and has no zero in a disk Iz - Zo I < 6 interior to the disk Izl < p. We are going to show that, in the disk Iz - Zo I < 6, we have log J(z)

1 211'

=-

121< log IJ(pe''P)I. . pei'P + z dIP pe''P - z

0

h

P - L-Iog ( ~=1 Pz ~

a~z

a~

)

2-b I'z L-Iog ( b) k

2-

+

~

p -

1'=1

p

Z -

I'

.

+ tC

,

(1.41)

where the logarithms are all holomorphic branches in the disk Iz - zol < 6 and e is a real constant. First note that the integral in (1.41) defines a function I(z) holomorphic for Izl < p. In fact, from

17

Nevanlinna's Theory of Meromorphic Functions

it follows that

Next note that both sides of (1.41) (the last term ic being excepted) are holomorphic functions in the disk Iz - Zo I < 6 and by (1.1) their real parts are equal, hence they differ only by a constant. Differentiating both sides of (1.41), we get,

(1.42)

This formula holds for Iz - zol < 6 and in particular for z = zoo Since Zo is arbitrary, (1.42) holds for Izl < p. (At a zero or pole of I(z) in the disk Izl < p, both sides of (1.42) become 00.)

Theorem 1.2. Let I(z) be a meromorphic function in a domain Izl < R (0 < R :5 00) such that Co = 1(0) =1= 0,00. Then for 0 < r < p < R we have m

(r, -I1') < 4log+ T(p,!) + 3log+ - -r + 4log+ p + 2Iog+-r 1 P-

+ 4log + log +

1

1

~

+ 16 .

( 1.43)

This theorem plays an important role in Nevanlinna's theory of meromorphic functions.

Proof. Consider a disk have

Izl < p (0 < P < R).

By (1.42), in this disk, we

18

Fix-points and Factorization of Merom orphic Functions

Next from

and

we have I

f'(z) 1 2p (1r7r i J(z) ~ (p -lzl)2 21r 10 Ilog IJ(pe 'P)lldcp

t

1p2 - a>.z 1

t

1p2 - b z I)

+ >.=1 p(z - a>.) + ,.=1 p(z - ;,.)

(1-44)

By (1-16),

1 21r

10r Ilog IJ(pei'P) Ildcp = m(p, J) + m 27r

(

1) ~ 2T(p, J) p, 7

1 + log ~

-

(1-45) On the other hand, if we set

p2 - zoz X(z,zo) = ( ) p z - Zo

(Izol <

p,Zo =I- 0) ,

(1-46)

then by (1-16),

m(r,x(z, zo))

+ N(r, X(z, zo)) = m (r, X(1 )) + N (r, (1 )) Z,Zo X z, Zo p + log ~ (0 < r < p) .

Since m

1)_ (( _ + r (r, X(1Z,Zo )) = 0, N (r, X(z,zo ) - 0, N r,X z, zo)) - log -I-I ' Zo

Nevanlinna's Theory of Meromorphic Functions

we get

19

P + r m ( r, X ( z, zo)) = log rz;;!-log rz;;!'

(1.47)

(1.44), (1.45) and (1.46) yield

J'(z)

I J(z) I ~

2p { (p-l zl)2 2T(p,J)

1

+log~ +

E h

k

Ix(z,a).)1 + : ; IX(z,b,.)1

}

In this inequality, setting z = re ilJ , taking log+ of both sides and making use of (1.22) and (1.23)' we find log

+

ilJ ) I 1J'(re + 2p J(re ilJ ) ~ log (p _ r)2

+ + + 1 + log 2T(p, J) + log log ~

h

+

L log+ Ix (re

k

ilJ ,

a).)1 +

L log+ Ix(re

ilJ ,

b,.)1

Next integrating both sides with respect to 0 from 0 to 211", multiplying both sides by 1/211" and making use of (1.47), we find m

( r, fJ')

~ log

+ (p _2pr)2 + log + 2T ( 1 p, ) J + log + log + ~

7) - N (r, 7) + N(p, J) - N(r, J) n (p, 7) n(p, J)

+ N (p, + log {

+

+ 2} .

(1.48)

For the 'sake of simplicity set

n(r) = n (r,

7) + n(r, J),

N(r) = N (r,

7) + N(r, J) = for n~t) dt . (0 <

r

Taking a value Pl such that 0 < r < PI < P < R and setting p (1.48), we obtain m

( r, fJ')

~ log

+

< R)

=

PI in

2PI + ( ) + + 1 (PI _ r)2 + log 2T Pl> J + log log ~

+ N(pt} - N(r)

+ log(n(PI) + 2) .

(1.49)

20

Fix-points and Factorization of Meromorphic Functions

In what follows, we are going to find upper bounds of n(pd and N(pd N(r). First from

n(pd log..E..Pl

~ [P n(t) dt ~ Jp1 t

N(p)

~ T(p, f) + T = 2T(p, f)

log -p Pl

-Pl) = -log (1 p - P

(p,

-f1)

1 + log ~ ,

P - Pl P

~--

we have

n(pd

~

_P- {2T(P, f) + log+ P - Pl

~ log _P- +

log+ n(Pl)

P - Pl

-Ill} , Co

log+ 2T(p, f) + log+ log+ _111 + log2 Co (1.50)

Next, N(r) being a convex function of log r, we have

N(pd - N(r)

~g~

~ -1 ~ {N(p)

og

log

- N(r)}

r

~g~

~ -1 ~ N(p)

og

r

e; {2T(p, f) + log + ~ 1 }

~ log;

which together with the inequalities

r)

Pl = log (1+ Pl -logr

r

r

r

P- r P

{2T(P, f) + log+

-Ill} .

~

Pl - , r

P log-

~--

yield

N(pd - N(r)

~ ~r PlP -- rr

Co

(1.51 )

In this inequality we only assume r < Pl < p. Taking in particular Pl - r r(p - r) {T(p, f) + log+ + 1} P + 1 ' - 2

(1.52)

N(pd - N(r) < 1 .

(1.53)

rtT

(1.51) becomes

Nevanlinna's Theory of Meromorphic Functions

21

It remains to find upper bounds of 1/(PI - r) and 1/(p- PI). First we have 1{ 2[T(p,f) +log+ - 1 } -1- = -1 + 1]p+ 1 , PI - r P- r r Icol 1 1 1 log+ - - ~ log+ - - + log+ - + log+ P + log+ T(p, f) PI - r P- r r 1

+ log+ log+ ~ + 2log 2 + log 3.

(1.54)

Next we have

P - PI = (p - r) - (PI - r) = (1 - A)(p - r) , where

A=

r

2{T(p, f) + log+

Vor +

1}p + 1

<~

2'

and hence P-PI

1

> 2(p-r),

1 1 log+ - - ~ log+ - - + log 2 . P - PI P- r

(1.55)

Finally from (1.49), (1.50), (1.53), (1.54) and (1.55), we get m

(r, -,1') < log+ PI + 2log+ - + log+ PI 1

r

+

~

i

T(PI'

f) + log+ log+ -I1-I Co

+ log+ n(PI) + 4log2

log+ PI + 2log+ _1_ + log _P- + 2log+ T(p, f) PI - r P - PI 1

+ 2log+ log+ ~ + 1 + 6log2 1

1

~ 4log+ P + 31og+ - - + 2log+ - + 4log+ p-r r

1 + 4log + log + ~ + 11 log 2 + 2log 3 + 1 .

Hence we have (1.43).

T(p, f)

Fix·points and Factorization of Merom orphic Functions

22

In Theorem 1.2, it is assumed 1(0) to, 00. IT the point z = 0 is a zero or pole of the function I(z), supposed non-identically equal to zero, then in the neighborhood of the point z = 0,

I(z) = c.z'

+ c.+1z·+1 +...

(c. to) .

We can apply Theorem 1.2 to the function JI(z) = z-' I(z) and get, for R, the inequality

o< r < p <

(r, 1') 11 < 4log+ T(p, Id + 3log+ - - + 4log+ P + 2log+ -

m

1

1

1

P- r

r

1 + 16 . + 4 log + log + j;J

To get an upper bound of m(r,

(1.56)

f), first from

If(z) JI(z) -

I'(z) I(z) ,

8

-----+-z

we have (1.57) Next from

m(r, Id

~

m(r, z-·)

+ m(r, J) ,

>0 - 8 log+ r if 8 < 0 , N(r,/d = N(r, I) - n(O, J) log r ,

m(r, z-.) =

where n(O, I) is equal to 0 or

T(r,

8

{

log + 1

if

8

r

-8, according to 8 > 0 or 8 < 0, we have

Id ~ T(r, J) + 181 (log+ r + log+ ;:)

(1.58)

(1.56), (1.57) and (1.58) yield m

(r, -I1') < 4log + T+ (p, J) + 3 log + - - + 8log + p + 6 log+ 1

1

p- r

r

1 + 4log+ log+ ~ + 5log+ 181 + 9log2 + 16,

Nevanlinna s Theory of Meromorphic Functions

23

where T+ (p, f) = max{T(p, I), o}. Evidently this inequality also holds, when 1(0) =1= 0,00. We have therefore the following corollary:

Corollary 1.1. Let I(z) be a meromorphic function non-identically equal to zero in a domain Izl < R (0 < R ~ 00). Then for 0 < r < p < R we have m

< 4log + T+ (p, f) + 3 log + - - + 8 log + P + 6log + (r, -1') 1 p-r r 1

1

1

+ 4log+ log+ ~ + 5log+ lsi + 25.

(1.59)

Sometimes we need the following generalization of Corollary 1.1:

Theorem 1.3. Let I(z) be a meromorphic function non-identically equal to zero and n 2: 1 a positive integer. Then there are positive constants A, B, C, D such that for 1 ~ r < p we have m

( rI(n)) 'j <

1

Alog+T(p,f) + Blogp+Clog+ p_r+D.

(1.60)

Proof. By Corollary 1.1, Theorem 1.3 holds when n = 1. Now suppose that Theorem 1.3 is true for a positive integer n. We are going to show that it is also true for n + 1. For this purpose, distinguish two cases: 10 I(n) (z) == o. In this case, obviously there are positive constants A1,B1,Cl, Dl such that for 1 ~ r < p we have m ( r,

I(n+l))

-1- <

A1log+ T(p, f)

+ Bl log P+ C 1 log+

1

p_ r

+ Dl

. (1.61)

2°/(n)(z) "t o. In this case, we first deduce from the identity I(n+l)(z) I(n)(z) I(n) (z) I(z) the inequality

I(n+l)) (/(n+l)) (/(n)) m ( r'--I- ~m r,-yN +m r ' j

(1.62)

Fix-points and Factorization of Meromorphic Functions

24

Next, by hypothesis, (1.60) holds and, on the other hand, by Corollary 1.1, we have, for 1 ::; r < p,

I(n+l)) 1 < 4log+ T(p, I(n)) + 8log+ P + 3log+ - - + d, (1.63) In p-r

m ( r, -(-)-

where d is a positive constant. Now let 1 ::; r < P and set PI = (r + p)/2, then m (r, I(n(+)l)) < 4log+ T(PI' I(n)) +8Iog+ PI +3Iog+ _1_ +d. (1.64) I n PI - r Since

T(PI' I(n))

=

m(PI' f(n)) + N(PI' f(n)) ::; m(pI' f) + m ( PI,

I(n)) T

+ (n+ I)N(PI,J) ::; (n + I)T(PI! J)

+m

( PI!

f(n)) T (1.65)

and by (1.60), m (PI!

f(/n )) < Alog+ T(p, J) + Blogp + Glog+ _1_ + D , p - PI

(1.66)

we see finally from (1.62), (1.60), (1.64), (1.65) and (1.66) that there are positive constants AI, B I , G I , DI such that (1.61) holds for 1 ::; r < p. In order to get some estimates of m(r, f(n) / f), which are convenient for certain applications, we need the following lemma due to Borel:

Lem.ma 1.1. Let cp(x) be a continuous non-decreasing function for x > 0, tending to 00 with x. Then the inequality II'

{x +

log~(x) }

< {cp(x)}2

(1.67)

holds, when x is exterior to a sequence of intervals of finite total length.

Proof. If there is a positive number a such that (1.67) holds for x ~ a, then the conclusion of Lemma 1.1 is evident. We may therefore assume that the inequality (1.68)

25

Nevanlinna's Theory of Meromorphic Functions

is satisfied by arbitrarily large values of x. Let Xo be a value satisfying cp(xo) > 1 and (1.68). Consider the sequence of values

xo, Xl Xn

1

= Xo + Iogcp (Xo) , x2 =

1

Xl

+ Iogcp (Xl) , ...

,

1

= Xn-l + Iog cp ( Xn-l ) , ...

(1.69)

.

Obviously xn(n = 0,1,2, ... ) is an increasing sequence. We are going to show that not all terms of this sequence satisfy (1.68). In fact, assume, on the contrary, that (1. 70) Then log cp(xn+d ~ 2 log cp(xn)

(n = 0, 1,2, ... )

(1.71)

and we have successively logcp(Xd ~ 2Iogcp(xo) , log cp(X2) ~ 22 log cp(xo) ,

which shows that cp(xn) tends to n-l

Xn

= Xo +

00

with n. On the the other hand,

1

L j=O log cp( Xj)

1

< Xo +

-

n-l 1

L --:- ,

log cp( xo) j=o 23

which shows that Xn is bounded. We get a contradiction. Let Xmo be the first term of the sequence xn(n = 0, 1,2, ... ) satisfying (1.67), obviously mo ~ 1. Set X = x mo ' then (1. 72) 1

X - Xo <

- log cp(xo)

mo-l

1

L --:- . j=o 23

(1.73)

26

Fix-points and Factorization of Meromorphic Functions

Since X satisfies (1.67), by continuity there is an interval (X, x~) in which (1.67) holds and such that x~ satisfies (1.68). By (1.72), (1.74) Starting from

X' =

X~l

>

x~

and repeating the same process, we get again a value

x~(ml ~ 1) satisfying (1.67) and such that

(1.75) 1

ml-l

1

X' - x'0< '" ~ . - log
(1. 76)

3=0

Since X' satisfies (1.67), there is an interval (X', x~) in which (1.67) holds and such that x~ satisfies (1.68). By (1.75), (1. 77) In this way we get successively a sequence of intervals

(xo, X), (x~, X'), (x~, X"), ... ,(x~p), X(p)), ...

(1. 78)

such that (1.67) holds in the intervals

[X,x') ,xo , ... , [X(p) , Xo(PH)) , ... o , [X''') and that we have in general

(1. 79)

(1.80) where mp

~

1. We then have successively log
, log
Nevanlinna's Theory of Meromorphic Functions

which implies that x~p) tends to

00

1

X- Xo

X' -

X'

< -

27

with p. Next we have

mo-l

1

L ---:-, log rp(Xo). 23 3=0

<

1

o - log rp(xo)

Consequently the total length of the sequence of intervals (1.78) does not exceed 1 ~ 1 2 logrp(xo)

f;

2i

=

logrp(xo) .

Corollary 1.2. Let J(z) be a non-constant meromorphic function and n;::: 1 a positive integer. Then the following estimates of m(r, J(n) / J) hold: 1° IT the order of J(z) is finite, then m

( rJ(n)) ' l = O(logr) .

(1.81)

2° In the general case, there is a sequence of intervals {Ip} of finite total length and depending only on J(z), such that when r is exterior to {Ip}, we have m

( rJ(n)) ' l =O{logT(r,f)+logr}.

( 1.82)

Proof. IT the order). of !(z) is finite, then

T(r, f) <

when r is sufficiently large. In (1.60) taking p (1.83), we see that when r is sufficiently large, m

(r, !(rn ))

where K is a positive constant.

(1.83)

rA+l ,

< Klogr,

=

2r and making use of

28

Fix-points and Factorization of Meromorphic Functions

In the general case, T( r, J) is continuous, non-decreasing for r > 0 and tends to 00 with r. We can then apply Lemma 1.1 to the function cp(r) = T(r, f). Consequently there is a sequence of intervals {Ip} of finite total length such that we have (1.84) when r is exterior to {Ip}. Let r be exterior to {Ip} and in (1.60) take p = r + 1/ log cp(r). Then making use of (1.84), we see that m

( r 'f(n)) T <

K'{log T(r, J)

+ logr}

,

when r is sufficiently large and exterior to {Ip}, where K' is a positive constant.

Remark. Consider a non-constant meromorphic function f(z), a finite value a and a positive integer n. By Theorem 1.3, for 1 ~ r < p, we have m

f(n)) ( r, -f< -a

A'log+ T(p,

1

f - a) + B'logp + G'log+ - - + D' , p-r

where A', B', G', and D' are positive constants. Since

T(p, f -

a)

~

T(p, J) + log+

lal + log 2 ,

we have also m (r,

:~)a)

< A'log+ T(p, J)

+ B'log p + G'log+

P

~ r + D".

(1.85)

On the basis of (1.85), evidently the two estimates (1.81) and (1.82) of m(r, f(n) / J) in Corollary 1.2 are also true for m(r, f(n) /(/ - a)). 1.6. SECOND FUNDAMENTAL THEOREM

Theorem 1.4. Let f(z) be a non-constant meromorphic function and ai(j = 1,2, ... ,qj q ~ 2) be q distinct finite values. Then for r > 0, we have

t m (r, f 3=1

~ a.) + m(r,J) ~ 2T(r,J) 3

Nt{r) + S(r) ,

(1.86)

29

Nevanlinna's Theory of Meromorphic Functions

where

Nd r) = {2N(r, J) - N(r, I')}

+N

(1.87)

(r, ;,)

and S (r) satisfies the following conditions: 10 IT the order of J(z) is finite, then

S(r) = O(log r) .

(1.88)

20 In the general case, there is a sequence of intervals {Ip} of finite total length and depending only on J(z), such that when r is exterior to {Ip}, we have S(r) = O{logT(r,J) +logr} . (1.89) This theorem is one form of the second fundamental theorem.

Proof. The method of proof is to introduce the auxiliary function 1

q

F(z) =

?= J(z) -

a.

1=1

1

and to find a lower bound and an upper bound of m(r, F). To find a lower bound of m(r, F), consider a value r> O. Set

0= min{l, laj - akl (1

~

i,k

~

q,i ¥: k)}

and define E j to be the set of values IP of the interval 0 the inequality

~

IP

~

21r, satisfying

. 0 IJ(re''P) - ajl < 2q .

IT z = re''P with IP E E j , then

F(z) =

1 {I + t;;."

J(z) - aj

J(z) - aj } J(z) - ak '

o 43 0

IJ(z) - akl ~ laj - akl-IJ(z) - ajl > 0 - 2q ~

" IJ(z) J(z) f;; IF(z)1 >

aj

I

2

2

ak < q 3q = 3" '

~I f(z)l_ aj

I,

(k

¥: i) ,

Fix-points and Factorization of Meromorphic Functions

30

hence log+

IF(rei~)1 ~ log+ I f( re'~. \

I-IOg3

- aj

(IP E E j )

-

Noting that the sets E j (j = 1,2, ... , q) are mutually disjoint, we have

m(r,F) ~ -1 Lq 211"

j=1

1 q ~211" " ~

1 1 E.

log+ IF(rei~)ldIP

J

E

j=1

log+

;

1

I/( re'~ .) -

d IO 3 a . I IP- g . J

Denoting by H j the complement of E j with respect to the interval 0

~

IP

~

211", then

-

11

211"

E;

log+

1

1

I(rei~)

- aj

dIP

1

=

(1) 1r-, aj

m

- -1

211"

log +1

r --

aj

1)

q ( m(r,F) ~Lm r, I-a. 3=1

H;

( , 1-1)

->m hence

1

1

I(rei~)

- aj

2q -logC '

2

-qlog cq -log3.

(1.90)

3

Now to find an upper bound of m(r,F), we write 1

F(z)

E q

= I'(z)

I'(z)

I(z) - aj

and obtain

m(r, F)

~ m (r, ;,) +

t,

m (r,

I~' aj) + log q .

Since m (r, ;,)

1dIP

= m(r, 1') + N(r, 1') - N (r, ;,) + log

I~I

Nevanlinna's Theory of Meromorphic Functions

31

by (1.16), where c =j; 0 is a constant, and

m(r, I') $ m(r, j) + m (r,

~)

,

it follows that

m(r, F) $ m(r, I) + N(r, 1') - N (r, ;,)

+m

(r, -1') + 2: m I

1') + log -I1+ log q . 1 c

(1.91)

2T(r, j) - Nl (r) + S(r) ,

(1.92)

q

(

r, _I. aJ

j=l

Inequalities (1.90) and (1.91) yield

~ m (r, I ~ aj) + m(r, j) $ where

S(r)

( f1') + ~

= m

q

r,

m

(

r,

1')

I _ a. + a

J=l

,

J

a being a constant. By Corollary 1.2 and Remark, evidently S(r) satisfies the conditions 10 and 2 0 in Theorem 1.4. Now we are going to state the second fundamental theorem in another form. For this, it is convenient to use the notations:

N(r, 00) = N(r,j),

N(r,a) = N (r, _ 1 )

I-a

(a finite)

(1.93)

introduced by Nevanlinna.

Theorem 1.5. Let I(z) be a non-constant meromorphic function and aj(j = 1,2, ... ,qj q 2: 3)q distinct values, finite or infinite. Then for r > 0, we have

q

(q - 2)T(r, j) $

2: N(r, aj) -

Ndr)

+ S(r) ,

(1.94)

j=l

where Ndr) is defined by (1.87) and S(r) satisfies the conditions 10 and 20 in Theorem 1.4.

Proof. We consider only the case that one of the values aj (j 1,2, ... ,q) is 00, for instance aq = 00. Then by Theorem 1.4, we have

I:

J=l

m (r,

I

~ a.) + m(r, j) $ 2T(r, j) J

Nl (r) + S(r) .

32

Fix-points and Factorization of Meromorphic Functions

To both sides of this inequality adding the sum Ej= 1 N (r, ai), and then noting that

T(r, f)

~T

(r, J ~

aJ

+ k i (j =

1,2, ... ,q - 1) ,

by (1.36), where ki(j = 1,2, ... ,q - 1) are positive constants, we get q

(q - 2)T(r,f) ~ LN(r,ai) - Ndr) + Sl(r) , i=l where Sl(r) = S(r) + I:j:~ ki evidently also satisfies the conditions 1° and 2° in Theorem 1.4. The case that the values ai(j = 1,2, ... ,q) are all finite, is treated by the same method. Now let us study the term Ndr) defined by (1.87). Ndr) consists of two parts 2N(r, f) - N(r, 1') and N(r, 1/1'). They are both non-negative for r ~ 1. Consider first the second part N(r, 1/1'). This part is related to the points Zo such that J(zo) is finite and Zo is a zero of order greater than one of the function J(z) - J(zo). For simplicity, let us name such a point Zo a multiple point of the first kind of J(z) and the order of Zo as a zero of J(z) - J(zo) the order of zoo Evidently n(t, 1/1') is equal to the number of multiple points of the first kind of J(z) in the disk Izl ~ t, each one of such points being counted as many times as its order minus one. Consider now the first part 2N(r, f) - N(r, 1'). This part is related to poles of order greater than one of J(z), namely multiple poles of J(z). Evidently 2n(t, f) - n(t, 1') is equal to the number of multiple poles of J(z) in the disk Izl ~ t, each multiple pole being counted as many times as its order minus one. Thus if we denote by n1 (t) the number of multiple points (those of the first kind and multiple poles) of J(z) in the disk Izl ~ t, each multiple point being counted as many times as its order minus one, then

ndt)

= {2n(t, f) - n(t, I')} + n (t, ;,)

(1.95)

Consequently we have the formula

Ndr)

=

l

o

r

n1(t) - ndO) t

dt+ndO)logr.

(1. 96)

Nevanlinna's Theory of Meromorphic Functions

33

The notation N(r, a) (a finite or infinite) introduced above may be expressed as

N(r,a) =

r n(t,a) -t n(O,a) dt+.n(O,a)logr,

10

(1.97)

where the meaning of n(t, a) is self-evident. Nevanlinna also introduced the following notation:

-

N(r, a) =

1" o

n(t, a) - n(O, a) dt + n(O, a) log r , t

(1.98)

where n(t, a) denotes the number of the roots in the disk Izl ~ t of the equation J(z) = a, each root being counted once. Noting that for any q distinct values, aj (j = 1,2, ... , q) finite or infinite, we have q

L

q

n(t, aj) - ndt) ~

j=1

L n(t, aj) , j=1

we deduce from Theorem 1.5 and (1.95) the following theorem: Theorem 1.6. Let J(z) be a non-constant meormorphic function and = 1,2, ... , qj q ~ 3)q distinct values finite or infinite. Then for r > 0 we have

ajU

q

(q - 2)T(r, 1) ~

L

N(r, aj) + S(r) ,

(1.99)

j=1 where S (r) satisfies the conditions 10 and 2 0 in Theorem 1.4. In what follows, we give some applications of the Theorems 1.4, 1.5 and 1.6. Corollary 1.3. Let J(z) be a transcendental meromorphic function. Then for each value a finite or infinite, the equation J(z) = a has an infinite number of roots, except for at most two exceptional values. This Corollary is Picard theorem for meromorphic functions. An exceptional value a, if it exists, is called a Picard exceptional value of J(z). Proof. In the particular case q = 3, Theorem 1.5 yields the inequality 3

T(r, 1) ~

L j=1

N(r, aj) + S(r) .

(1.100)

Fix-points and Factorization of Meromorphic Functions

34

Now suppose that for three values aiU = 1,2,3), the equations j(z) aiU = 1,2,3) all have at most a finite number of roots. Then evidently

N(r,ai)=O(logr)

U=1,2,3).

Consequently by the condition 2° in Theorem 1.4, when r is exterior to a sequence of intervals {Ip} of finite total length and is sufficiently large, we have T(r,j) ~ K{logT(r,f) +logr}, where K is a positive constant. But this is impossible, by (1.38). Corollary 1.4. Let j(z) be a transcendental meromorphic function of finite positive order p (0 < P < 00). Then for each value a finite or infinite, we have log n(r, a) 1-;(1.101) 1m = p, r-+oo log r except for at most two exceptional values. This Corollary is Borel's theorem for meromorphic functions. An exceptional value a, if it exists, is called a Borel exceptional value of j(z). Proof. Noting first that, for r

n(r, a) log2 ~

~

1

2r

r

1, we have

n(t a) -t-'-dt ~ N(2r, a) ,

and then by the first fundamental theorem,

n(r, a) log 2

~

T(2r, f) + k ,

where k is a positive constant. So we have

-1. log n(r, a) 1m < p. log r -

r-+oo

(1.102)

Now suppose that there are three values aiU = 1,2,3) which do not satisfy (1.101). Then we can find a constant )(0 < ) < p) such that

n(r,ai) < r'"

(r ~ ro)

Hence

N(r,ai)-N(ro,ai) =

l

r

ro

n(t,ai) dt< t

U = 1,2,3) .

l

r

ro

t'" 1 '" -ro). '" -dt=;:(r t

(1.103)

Nevanlinna's Theory of Meromorphic Functions

35

From (1.100), (1.103) and the relation S(r) = O(log r), it follows that, for sufficiently large values of r,

T(r, f) < hr>' , where h is a positive constant. But this is impossible because A < p. Consider a transcendental meromorphic function J(z) and a finite value a. By (1.36) we may write

Dividing both sides of this equality by T(r, f) and then taking lower limit, we get . m hm

r~

(r, J~a) T(r, f)

=

1-

_. N (r, J~a) hm ---'-:------,----'r-+oo

T(r, f)

(1.104)

We have also lim m(r, f) r~

=

1 _ lim N(r, f) .

T(r, f)

r-+oo

(1.105)

T(r, f)

Nevanlinna introduced the notation 6(a, f) defined for a finite or infinite as follows:

C()

- . N(r, a)

a, J = 1- r~~ T(r, f)

o

(1.106)

Evidently

o~

6(a, f) ~ 1 .

Corollary 1.5. Let J(z) be a transcendental meromorphic function and aj{j = 1,2, ... , qj q ~ 2)q distinct finite values. Then q

L 6(aj, f) + 6(00, f) ~ 2 .

(1.107)

j=l

Proof. By (1.86), we have

t - l - m (r _1_) j=l T(r, f) , J - aj

+ m(r,f) < 2+ ~ T(r, f) -

T(r, f) .

36

Fix-points and Factorization of Meromorphic Functions

Taking lower limit and making use of (1.104) and (1.105), we get

f; 6(a;, f) + q

S(~

•

6(00, f)

~ 2 + r~~ T(r, f) .

Since S(r) satisfies the condition 2° in Theorem 1.4, we have

lim~<

r-+oo

T(r, f) -

lim r-+oo

r/j! {I p}

~=O T(r, f)

and (1.107). IT a value a is such that 6(a, f) > 0, then a is called a deficient value of J(z) or a Nevanlinna exceptional value of J(z), and 6(a, f) the deficiency corresponding to the value a. It is easy to see that J(z) can have at most countably many deficient values. In fact if we denote by ak the set of the deficient values satisfying the inequality 1/(k + 1) < 6(a, f) ~ 11k, and by a the set of all the deficient values, then 00

a=

Uak k=1

and, by (1.107), ak consists of at most 2(k

+ 1)

deficient values. We have

L6(a,f) ~ 2,

(1.108)

a

where the summation is taken with respect to all the deficient values of J(z). In particular for a transcendental entire function J(z), we have 6(00, f) = 1 and hence (1.109) 6(a, f) ~ 1 .

L

a;too

Now we introduce another important notion that of completely multiple value. Consider a transcendental meromorphic function J(z). A finite value a is called a completely multiple value of f(z), if each zero of J(z) - a has an order greater than one. 00 is called a completely multiple value of J(z), if each pole of f(z) has an order greater than one.

Corollary 1.6. Let J(z) be a transcendental meromorphic function. Then J(z) has at most four completely multiple values. Proof. Assume that J(z) has five completely multiple values ai(j 1,2,3,4,5). In Theorem 1.6 taking q = 5, we get 5

3T(r, f) ~

L ;=1

N(r, ail + S(r) .

=

Nevanlinna's Theory of Meromorphic Functions

Since aj is completely multiple, we see that for r -

N(r, aj) ::;

1

2N(r, aj)

1

::; 2T(r, J)

~

37

1,

+ h, (j = 1,2,3,4,5)

where h is a positive constant. Consequently 1

2T(r, j) ::; 5h + S(r) which leads to a contradiction in taking account of the condition 2° in Theorem 1.4. In particular if J(z) is a transcendental entire function, then J(z) has at most two finite completely multiple values. In fact, if J(z) has three finite completely multiple values aj(j = 1,2,3), then in Theorem 1.6, taking q = 4, a4 = 00, we get 3

2T(r, j) ::;

E N(r, aj) + S(r) j=1

which also leads to a contradiction. After more than fifty years later and by following earlier work of C. Chuang, Frank-Weissenborn and C. Osgood, N. Steinmetz have now been able to present a most convincing proof of the following generalized result which was raised by Nevanlinna in 1929.

Theorem. (Nevanlinna's second fundamental theorem for small functions). Let J(z) be a transcendental meromorphic function and a1(z), a2 (z), . .. ,aq (z) be distinct q(~ 2) meromorphic small functions (including constant function) satisfying

T(r, ai(z)) = S(r, j)

as

r

-+

00

(i = 1,2, ... ,q) .

Then, for any e > 0,

t

m (r,

Corollary.

J ~ ai) + m(r, J) ::; (2 + e)T(r, j) + S(r, j) .

E 6(a(z), j) + 6(00, j) ::; 2 , a(z)

Fix-points and Factorization of Meromorphic Functions

38

where the summation is over all deficient functions of f including constants; a(z) is called a deficient function if T(r,a(z)) = S(r, J) and 1 -

r~~N (r, f _la(z)) /T(r,J) > o.

The basic ingredient of the proof on the above result is the success in replacing the lemma of logarithmic derivative (Theorem 1.2) by an estimation of m(r, P(J)I fh), P(J) a differential polynomial of f and h a positive integer.

Rem.ark. It is natural and interesting to find some non-trivial applications of the generalized results in the studies of fix-points and factorization theory of merom orphic functions. 1.7. SYSTEMS OF MEROMORPHIC FUNCTIONS In this paragraph our main purpose is to give a complete proof of the following theorem of Borel on systems of entire functions:

Theorem. 1.1. Let fj(z)(i

(n

=

1,2, ... , n) and gj(z)(i

=

1,2, ... , n)

~ 2) be two systems of entire functions satisfying the following condi-

tions:

1) ~;'=l Ij(z)egj(z) == O. 2) For 1 :::; i :::; n, 1 :::; h, k :::; n, hi- k, the order of Ij (z) is less than the order of eg,,(Z)-gk(Z): p(Jj) < p(eg,,-gk). Then IJ(z) == 0 (i = 1,2, ... ,n). This theorem has important applications in the theory of fix points and factorization of meromorphic functions. It gives rise to the research of Nevanlinna on systems of merom orphic functions. His main result is the following theorem:

Theorem. 1.S. Let !pj(z)(i = 1,2, ... ,n) be n linearly independent meromorphic functions satisfying the identity !PI

Then for 1 :::;

i :::;

T(r, !Pj) :::;

= 1.

(1.110)

+ N(r,!pj) + N(r, D) + S(r) ,

(1.111)

+ !P2 + ... + !Pn

n, we have

t

k=l

N (r,

~) !Pk

39

Nevanlinna's Theory of Meromorphic Functions

where !Pl

!P2

!P~

!P~

!Pn

!P~ (1.112)

D= (n-l) !Pl

(n-l)

!P2

...

(n-l)

!pn

and when r is exterior to a sequence of intervals {Jp } of finite total length,

S(r) = O{logT(r)

+ logr}

,

(1.113)

where

T(r) =

m~x

l~J~n

T(r, !Pi) .

(1.114)

Proof. Differentiating (1.110) successively, we get (k) !Pl(k) +!P2

+ ... +!Pn(k)

(k - 1, 2 , ... , n - 1) .

-- 0

(1.115)

Since !Pi(j = 1,2, ... ,n) are linearly independent, D is not identically equal to zero and, by (1.110) and (1.115), we have D

=

Di

(j = 1,2, ... ,n) ,

(1.116)

where Di is the minor corresponding to !Pi in D. Hence (1.117)

(1.118)

!Pl

!P2

!Pn

and ~l is the minor corresponding to the element 1 in the first row and the first column of ~. From (1.117), we get

m(r,!pd S;

m(r'~l) + m (r, ~)

S;

m(r,~d + m(r,~) + N(r,~) + h, (1.119)

Fix-points and Factorization of Meromorphic Functions

40

where h is a constant_ Next from

D

!:l.=---IPI IP2 ••• IPn

we get

N(r,!:l.)

~ N(r, D) + t

N (r,

j=1

~)

(1.120)

IPJ

On the other hand, if we set

Sdr)

m(r, !:l.1)

=

+ m(r,!:l.) + h ,

then by (1.118) and Corollary 1.2, it is easy to see that there is a sequence of intervals {Jp } depending only on IPj(j = 1,2, ... n) and of finite total length, such that when r is exterior to {Jp }, we have

Sdr) = O{log T(r) + log r} .

(1.121)

Consequently (1.122) Similarly we have

T(r,IPj)

~

t

k=1

N

(r,~) + N(r, D) + N(r,IPj) + Sj(r) IPk

(1.123)

(j = 2,3, ... , n) , where Sj(r) is such that, when r is exterior to {Jp }, we have

Sj(r) = O{logT(r)

+ logr} (j = 2,3, ... ,n) .

(1.124)

Finally defining

S(r) = m!IX Sj(r) /$;J$;n

we get (1.111) and (1.113) from (1.122), (1.123), (1.121) and (1.124). Theorem 1.9. Let fj(z)(j = 1,2, ... , nj n 2: 2) be n merom orphic functions satisfying the following conditions:

41

Nevanlinna's Theory of Meromorphic Functions

1°

2:i=1 Cj Ij (z)

2°/j(z)

~ 0

== 0, where Cj(j = 1,2, ... ,n) are constants. (j = 1,2, ... ,n) and for 1 ~ j,k ~ n,j"# k,/j(z)/lk(z) is

not a rational function.

3° N(r, Ij) = o{r(r)}, N(r, 1/lj) = o{(r(r)}(j = 1,2, ... ,n), where r(r) = n:tin I~"k~n jt:-k Then

Cj

=

0 (j

=

Jk

1,2, ... , n).

Proof. Consider first the case n

Assume that Cj(j then

{T (r, ~j) }

=

=

2. Then

1,2) are not both equal to zero, for example

C1

"#

0,

h(z) _ Cz h(z) =-~

which is incompatible with condition 2°. Consequently Theorem 1.9 holds when n = 2. Now assume that Theorem 1.9 holds for an integer n ~ 2, and let us show that it is also true for n + 1. In fact, consider n + 1 meromorphic functions Ij(z)(j = 1,2, ... ,n+l) satisfying the conditions in Theorem 1.9, so that n+1

L

cj/j(z) == 0 .

(1.125)

j=1

Suppose that Cj(j = 1,2, ... ,n + 1) are not all equal to zero. Then Cj(j = 1,2, ... ,n + 1) must be all different from zero. In fact, if for example Cn +1 = 0, then n

L cj/j(z) == 0 j=1

and Ij(z)(j = 1,2, ... ,n) satisfy the conditions in Theorem 1.9. Since by assumption Theorem 1.9 holds for the integer n, we have Cj = 0 (j = 1,2, ... ,n) and hence Cj = 0 (j = 1,2, ... ,n + 1), contrary to the hypothesis that Cj (j = 1,2, ... ,n + 1) are not all equa1 to zero. So Cj "# 0 (j = 1,2, ... ,n + 1). Set

lr>,'(Z) =_ r

c;lj(z)( )

cn +1/n+1

Z

(J, = 1,2, ... ,n ) .

(1.126)

42

Fix-points and Factorization of Meromorphic Functions

By (1.125)' we have n

L
Lai
n

L aicifi(z) == 0 . i=1 Since, by assumption, Theorem 1.9 holds for the integer n, we have aici = 1,2, ... , n). On the other hand, at least one of ai(j = 1,2, ... , n) is different from zero, for instance a1 "I 0, so C1 = 0 contrary to the result Ci "I O(j = 1,2, ... , n + 1) obtained above. It follows that we can apply Theorem 1.8 to the functions
o (j =

T(r,
t

k=1

N (r,

....!....) + N(r,
(1.127)

(j = 1,2, ... , n) . By the condition 3° in Theorem 1.9, it is easy to see that N(r,
T(r,
(1.128)

where S(r) satisfies (1.113) when r ~ {Jp }. But (1.128) is impossible, because by the condition 2° for fi(j = 1,2, ... , n + 1) in Theorem 1.9 and by (1.126),
lim T(r) = r-+oo

log r

00 •

43

Nevanlinna's Theory of Meromorphic Functions

This contradiction proves that the coefficients Cj(j (1.125) are all equal to zero, as we want to show.

=

1,2, ... , n

+ 1)

in

Theorem 1.10. Let fj(z) (j = 1,2, ... , n) and gj(z) (j = 1,2, ... , n) 2) be two systems of entire functions satisfying the following conditions: 1) 2:;=1 fj(z)egj(z) = O. 2) For 1:::; j, k:::; n,j =I k, gj(z) - gk(Z) is non-constant. 3) For 1 :::; J' :::; n, 1 :::; h, k :::; n, h =I k,

(n

~

(1.129) Then fj(z)

=0

(j = 1,2, ... , n).

Proof. Consider first the case n

If Ii (z)(j O. Then

=

=

2. Then

1,2) are not both identically equal to zero, for instance

II (z) 1=

and by the condition 3), we get

T(r,e g1 - g2 ) = T(r,h/fd :::; T(r, h) + T(r, l/fd = T(r, h) + T(r, fd + a = o{T(r, eg1 - g2 )} , which is impossible. Now assume that Theorem 1.10 holds for an integer n ~ 2. To show that it also holds for n + 1, let fj(z), gj(z) (j = 1,2, ... , n + 1) satisfy the conditions in Theorem 1.10, and assume that fj(z) (j = 1,2, ... , n + 1) are not all identically equal to zero. Then we must have

fj(z) 1= 0 In fact, if for instance fn+l (z)

(j

=

1,2, ... ,n+ 1).

= 0, then

n

L j=1

fj(z)egj(z)

=0

44

Fix-points and Factorization of Meromorphic Functions

and fi(z) (j = 1,2, ... , n) satisfy the conditions in Theorem 1.10, consequently, by assumption, J,-(z) == 0 (j = 1,2, ... , n) and hence fi(z) o U = 1,2, ... , n + 1) contrary to the hypothesis. Now set

Fj(z)

= fj(z)eUj(z),

Cj

= 1 U = 1,2, ...

, n + 1)

(1.130)

then by condition 1) n+1

L

cjFj(z) == 0 .

(1.131)

j=1

Obviously Fj(z) :f:- 0 U = 1,2, ... , n + 1) and, Theorem 1.10 being true for n = 2, when 1 ~ j,k ~ n+ I,J- t= k,Fj(z)/Fk(Z) is not a rational function. Moreover,

N(r,Fj)=O and, for h

t= k

(1 ~ h, k ~ n

+ 1)

U=1,2,oo.,n+1) we have

So F, (z) U = 1,2, ... , n+ 1) satisfy the conditions 1°,2°,3° in Theorem 1.9 and consequently Cj = 0 U = 1,2, ... , n + 1) contrary to (1.130). This proves Theorem 1.10. Finally for the proof of Theorem 1. 7, we still need the following two lemmas:

Lemma 1.2. Let 9(z )

=

CoZ

m.

+ CIZ m.-l + ... + Cm.

(m ~ l,eo

t= 0)

Nevanlinna's Theory of Meromorphic Functions

45

be a polynomial and define

A(r, g) = max Re{g(z)} . Izl=r

Then when r is sufficiently large, we have (1.132)

Proof. Set

z = re i8 ,

Cj

=

ICjle iaJ

(j = 0,1,2, ... m) .

Then

+ ICllrm-lei((m-l)8+ad + ... + Icmleiam , = Icolrm cos(m8 + 00) + ICllrm-1 cos((m - 1)8 + od + ... + Icml cos Om . (1.133)

g(re i8 ) = Icolr m ei (m8+ a o) Re{g(re i8 )}

In particular taking 8

=

80

=

-oo/m, we get

Re{g(rei80 )} = Icolrm + ICllrm-1 cos((m - 1)80 + 01)

+ ... + Icml cosOm . Since

Re{g(re i80 )}

::;

A(r,g) ,

obviously the first part of (1.132) holds. By (1.133), the second part of (1.132) also holds.

Lemma 1.3. Let I(z) be a transcendental entire function and set e/(z). Then

F(z) =

lim log log M(r, F) log r

=

00 •

r-+oo

Proof. We have M(r, F)

M(r, I) ::;

= eA(r,J)

and the inequality

R2~ r {A(R, j) + 21/(0) I}

(0 < r < R) .

(1.134)

46

Fix-points and Factorization of Meromorphic Functions

(See, for instance, Lemma 1.2 in the book Singular Directions of Meromorphic Functions.) In particular, taking R = 2r, we get

M(r, f) :::; 4{A(2r, f)

+ 2If(0)1}

.

On the other hand, by (1.39), we have lim log M(r, f) log r

= 00

lim log A(r, f) log r

= 00

•

r-+oo

Hence r-+oo

and (1.134) follows. Now let us prove Theorem 1.7 as follows: By Theorem 1.10, it is sufficient to show that the condition 2) in Theorem 1.7 implies the conditions 2) and 3) in Theorem 1.10. In fact, since the order of an entire function is always non-negative, hence for h '1= k, the order of egh(Z)-gk(Z) is greater than zero and gh(Z) - gk(Z) is non-constant. So the condition 2) in Theorem 1.10 is satisfied. To see the condition 3) in Theorem 1.10 is also satisfied, consider a function fj(z) and two functions gh(Z), gk(z)(h '1= k). Since the order of J,-(z) is less than that of egh(Z)-gk(Z), the order p of fj(z) is finite. Distinguish two cases: A. gh(Z) - gk(Z) is a polynomial of degree m ~ 1. In this case, by Lemma 1.2, evidently the order of egh(Z)-gk(Z) is m. Hence p < m. Taking a number). such that p < ). < m, then when r is sufficiently large,

and by Lemma 1.2, (1.135) On the other hand, by (1.33),

Consequently, (1.136)

Nevanlinna's Theory of Meromorphic Functions

47

B. gh(Z) - gk(Z) is a transcendental entire function. In this case, by Lemma 1.3, . log log M(r, eUA -Uk) hm = 00. r-oo logr Taking two numbers A, A' such that p < A < A' when r is sufficiently large, we have log M(r, eUA-Uk) > r A ' • It follows that (1.135) and (1.136) are also true. So condition 3) in Theorem 1.10 is also satisfied. Theorem 1.7 is now completely proved.

Corollary 1.'1. Let fi(z)(i = 1,2, ... ,n+l) and gi(z)(j = 1,2, ... ,n) (n ~ 1) be two systems of entire functions satisfying the following conditions:

Ei=l

fi(z)eUj(z) == fn+1(z). 2) For 1 ~ j ~ n+ 1, 1 ~ h ~ n, the order of fi(z) is less than the order of eUA(z). In case n ~ 2, for 1 ~ j ~ n + 1, 1 ~ h, k ~ n, h ¥= k, the order of fi(z) is less than the order of eU"(Z)-Uk(Z).

1)

Then fi(z)

== 0 (i = 1,2, ... ,n+ 1).

Proof. It is sufficient to note that, in setting gn+dz) = 0, we have n

L

fi(z)eUj(z) - fn+dz)eu,,+.(z)

i=l

and then Corollary 1. 7 follows from Theorem 1. 7.

== 0

,

2 FIX-POINTS OF MEROMORPHIC FUNCTIONS

2.1. INTRODUCTION In the theory of fix-points of meromorphic functions, Nevanlinna's theory of meromorphic functions and Montel's theory of normal families play an important role. So in this chapter we first give some complements of Nevanlinna's theory of meromorphic functions sketched out in the previous chapter and prove some theorems on fix-points due to Rosenbloom and Baker. Next we give an account of the main points of Montel's theory of normal families of holomorphic functions and apply it to Fatou's theory of fix-points of entire functions. 2.2. SOME THEOREMS ON MEROMORPHIC FUNCTIONS We first prove the following theorem which is a generalization of Nevanlinna's second fundamental theorem.

Theorem 2.1. Let I(z) be a non-constant meromorphic function. Let = 1,2,3) be three distinct meromorphic functions such that


T(r,
In

(j = 1,2,3) .

(2.1)

Then for r > 1 we have the inequality

T(r, f)

$

t.

N (r, I _1
+ o{T(r,

where S(r) satisfies the following conditions: 49

In + S(r)

(2.2)

Fix-points and Factorization of Meromorphic Functions

50

1° IT J(z) is of finite order, then

S(r) = O(log r) 2° In general, there exists a sequence of intervals {Ip} of finite total length such that for r ~ {Ip} we have

S(r) = O{log T(r, f) + log r} . Proof. Consider the auxiliary function

F(z) = J(z) - cPt{z) CP2(Z) - CP3(Z) J(z) - CP3(Z) cp2(Z) - cPt{z)

(2.3)

We have

where .A is a constant. Then by the identity 1 + CP3 - CP1 J - CP3

=

CP2 - CP1 F , CP2 - CP3

we get

$.

T(r, CP3 -1 CP1 ) + T(r, CP3CP2 -- CP1CP3 F) + log 2

$. T (r,

1 ) CP3 - CP1

+ T (r,

CP2 - CP1) CP2 - CP3

+ T( r, F) + log 2

.

(2.4)

From (2.3) and (2.4) we deduce 3

T(r, f) $. 3 LT(r, cPj) + T(r, F) + .A1

(2.5)

j=l

where .A1 is a constant. (2.5) shows that the function F(z) is non-constant, consequently by Theorem 1.6, we have, for r > 0,

T(r, F) $. N (r,

~)

+ N (r, F ~

1) + N(r,F) + S(r)

(2.6)

51

Fix-points of Merom orph ic Functions

where S(r) satisfies the conditions 1° and 2° in Theorem 1.4 with respect to F(z). We have

N(r, F) ~ N (r, 1- IPI) + N I - IP3

= N (r, 1 +

IP2 - IP3) IP2 - IPI

IPI) + N (r,

IP3 -

I-

(r,

IP3

IP2 - IP3 ) IP2 - IPI

-(r'-I-1) +N(r,IP3-IPd+N -(

~N

- IP3

r,

IP2-IP3) IP2 - IPI

Since

N (r,

IP2 - IP3) IP2 - IPI

~ T (r,

IP2 - IP3) IP2 - IPI

we get

~ N (r, I ~ IP3)

N(r, F)

+ o{T(r,

In .

(2.7)

Similarly we get

N (r,

~ ) ~ N (r, 1_1 IPI) + o{T(r, J)} ,

N (r, F ~

1) ~ N (r, I ~ IP2) + o{T(r, In .

(2.8)

(2.9)

Inequalities (2.5)-(2.9) yield

T(r,J)

~ t N (r, I~ i=1

It remains to show that rem 2.1. In fact we have

S(r)

.) +o{T(r,In+S(r). IPJ

satisfies the conditions 1° and 2° in Theo-

T(r, F) ~ T (r, 1- IPI) + T (r, I - IP3 and by the identity

I I -

IPI IP3

IP2 - IP3) IP2 - IP 1

IP3 - IPI I - IP3 '

---=1+':"'--='-':""'="

Fix-points and Factorization of Merom orphic Functions

52

we have

3

T(r, F) ~ T(r, f) + 3 LT(r,!Pi) + >'2 i=l where >'2 is a constant. Now it is evident th~t S(r) also satisfies the conditions 10 and 2° in Theorem 2.1. Similarly we can prove the following theorem:

Theorem 2.2. Let I(z) be a non-constant meromorphic function. Let !Pi (z)(i = 1,2) be two distinct meromorphic functions such that

T(r, !Pi)

= o{T(r, In

(j

= 1,2)

.

Then for r > 1 we have the inequality

T(r, f)

~ N(r, f) +

t.

~ !Pi) + o{T(r, In + S(r)

N (r, I

where S(r) satisfies the conditions 1° and 2° in Theorem 2.1. Now we are going to prove some theorems on the growth of composite functions. For this we need some preliminary lemmas and theorems.

Lemma 2.1. Let a and b be two positive numbers such that b ~ 8a 2 • Then for x ~ 2, we have ~

+ 8b

e 4 x > 8a log x

(2.10)

.

Proof. Consider the auxiliary function b.

!p(x) = e x - 8a log x - 8b . 4

Evidently it is sufficient to show that

!p(2) > 0,

!p' (x) = e

8a

b. 4

-

-

x

> 0 (x

~ 2) .

(2.11)

We have 2e!

>

2(1 +

!!.a

+ .!:. b22 ) >

1+ ( 1+

~)

2

(2.12)

2a

b a

- >

8a,

> 8a ( 1 +

~)

> 8alog2 + 8b

= 8a

+ 8b

(2.13)

Fix'points of Meromorphic Functions

53

and hence ~

2e .. > 8alog2

+ 8b

which shows that the first inequality in (2.11) holds. On the other hand, (2.12) and (2.13) imply ~ 2e .. > 8a , hence the second inequality in (2.11) also holds.

Lemma 2.2. Let U(r) be a non-negative and non-decreasing function in an interval 0 < r < p. Let a and b be two positive numbers such that b ~ 2a and b ~ 8a 2 • Assume that the inequality

R

U(r) < alog+ U(R) + alog R _ r + b

(2.14)

holds for 0 < r < R < p. Then the inequality

R

U(r) < 2a log R _ r + 2b holds for 0 < r < R < p. This Lemma in a different form was obtained by Borel in his fundamental paper "Sur les zeros des fonctions entieres", Acta Math. 20 (1897). It is put in the present form by Bureau and Milloux.

Proof. Let rand R be two values such that 0 < r < R < p and assume

R

U(r) ~ 2alog R-r +2b.

(2.15)

We are going to show that the two values r' = (r + R)/2 and R also satisfy the inequality

U(r')

~ 2alog R ~ r' + 2b .

In fact, by (2.14)' we have

U(r) <

a log+

r' r' - r

U(r') + a log - - + b

R < alog+ U(r') + a log - - + b • R- r' From this inequality and (2.15) written in the form

R

U(r) ~ 2alog R _ r' - 2alog2 + 2b ,

(2.16)

Fix·points and Factorization of Meromorphic Functions

54

we get

R b log+ U(r') > log R _ r' - 210g2 + ~ that is log+ U(r') > log

,

(~e! R ~ r,)

Since b ~ 2a, we have 1k R 1k 12 > -eo. > -e > 1 4 R-r' 4 - 4 ' ') 1 k R U (r > - e " - - . 4 R-r'

-eo. - -

On the other hand, by Lemma 2.1, taking x have k

(2.17)

= RI(R - r') in (2.10), we

R R > 8a log - - + 8b . R-r' R-r'

(2.18)

eo. - -

Inequalities (2.17) and (2.18) yield (2.16). Now let rn

=

(rn-l

+ R)/2 (n = 1,2, ... ),

ro

=

r.

Then for each n, we have

0< rn < R < P and

R U(rn) ~ 2alog R _ rn

(2.19)

+ 2b .

(2.20)

U(r) being a non-decreasing function, (2.19) implies U(rn) ~ U(R). On the other hand, (2.20) implies U(rn) -+ 00, as n -+ 00. So we get a contradiction. Lemma 2.3. Let a > e and x > 0 be two numbers. Then we have 1 logx + alog+ log+ - ~ a(loga - 1) x

+ log+ x

.

(2.21)

Proof. If x > lie, then log+ 10g+(1lx) = 0, hence (2.21) holds. If (2.21) becomes

x ~ lie, then

log x

+ a log log -1 x

~

a(log a - 1) .

(2.22)

55

Fix-points of Meromorphic Functions

Consider the function

rp(y) = alogy - y - a(loga -1)

(y > 0)

(2.23)

and its derivative

rp'(y)

=

~

- 1 .

(2.24)

Y

We can see easily

rp(y)

:::=;

rp(a) = 0 (y > 0) .

(2.25)

Replacing in (2.25) y by log(l/x), we get (2.22).

Theorem 2.3. Let I(z) be a holomorphic function in the circle Izl < 1 in which I(z) does not take the values 0 and 1. Then for Izl < 1, we have log I/(z)1 <

1~

Izl

(A log+ 1/(0)1 + Blog

1:

(2.26)

Izl)

where A and B are two positive numerical constants. This is the classical theorem of Schottky.

Proof. Consider the auxiliary function 1

F(z)

1

I(z) + I(z) - 1

=

As in the proof of Theorem 1.4, we see that for 0 < r < 1, we have

m(r, F)

~m

(r, 7-)

+ m

(r, I

~

J-

2log 4 -log 3

and

1') +loglf'(0)I+log2 1 ( 1') +m (r'/_1

m(r,F):::=;m(r,J)+2m r,! where we assume 1'(0) m

(r, 7-)

+ m

=I o. Hence

(r, I

~

J

:::=;

m(r, J) + m

+ 2m

(r,

~)

1 (r, I 1') _ 1 + log 1I' (0) 1 +

a

56

Fix·points and Factorization of Meromorphic Functions

where a is a positive numerical constant. Then by (1.14), we have

m(r, f) = m (r,

-1 ) + log 1/(0)1 , ~ 1) + log 1/(0) -

m(r,1 -1) = m (r, I

11

and get

m(r, f)

~ 2m (r, ~) + m (r, I ~ 1) + log 1/(0)1 + log 1/(0) I

1

11

,

+ og 1/'(0)1 + a . Next by Theorem 1.2, for 0 < r < R < 1, we have m

( 7f) < r,

4log+ m(R, f) + 3 log R 1_ r 1 + + 1 + 2 log ;:- + 4log log 1/(0) I + 16

m

f) <4Iog+m(R,f)+3Iog R _ r ( 1-1 1

r,

+

1

+2Iog;:-+4Iog log

+

1

1/(0)_11+16.

Hence 1

m(r, f) < 121og+ m(R, f) + 9log R _

r

1

1

+ 6log ;:- + 8 log + log +

+

+ 4log log

+

II (0) I + log II (0) I

1

1/(0) _ 11

+ log 1/(0) - 11 + log

I/'~O)I + a" .

(2.27)

By Lemma 2.3, 1

8log+ log+ 1/(0)1 + log 1/(0)1 ~ 8(log 8 - 1) + log+ 1/(0)1, 1

I+

4log+ log+ 1/(0) _ 1

(2.28)

log 1/(0) - 11 ~ 4(log4 - 1) + log+ 1/(0)1 + log 2 . (2.29)

Fix-points of Meromorphic Functions

57

From (2.27), (2.28) and (2.29), we get the following result: For 1/2 ~ r < R < 1, we have

R

1

m(r, f) < 121og+ m(R, f) + 9log R _ 1 + 2log+ 1/(0)1 + log 1/'(0)1 + f3 where f3 is a positive numerical constant. In the interval 0 < r < 1 define a function U(r) as follows:

(o
U(r)=O

,

(~~r
U(r)=m(r,f)

Then U(r) is non-negative and non-decreasing for 0 < r < 1, and satisfies, for 0 < r < R < 1, the inequality

R U(r) < alog+ U(R) + alog-- + b R-

where

a

=

12,

b = 2log+

1/(0)1 + log+

r

1f'~0)1 + (3'

with (3' = max((3, 8a 2 ). Consequently by Lemma 2.2, for 0 < r < R < 1, we have R U(r) < 2a log R _ r + 2b . Keeping r fixed and letting R

U(r) Hence for 1/2

~

-+

~

1, we get 1 1-r

2a log - - + 2b .

r < 1,

m(r, J)

~ 2a log

1 -1-r

+ 2b

and then for 0 < r < 1, 1 1-r

m(r, J) < 2a log - - + 2b' with b'

= a log 2 + b. By (1.33), in taking R = 1, P = (1 + r)/2, we have

4 (1-+- , I

log+ M(r, J) ~ - - m 1- r

r

2

)

(0 <

r

< 1) .

Fix-points and Factorization of Meromorphic Functions

58

Hence for 0 < r < 1, we have 4 ( 2alog 1-r 2 +4Iog+ 1/(0)1+2Iog+ 1/'(0)1 1 log+ M(r,j) < 1-r +b 1) (2_30) where b1 is a positive numerical constant_ Now let us distinguish two cases: 10 II' (0) I ~ 1. In this case, by (2.30)' we have for 0 < r < 1, log M(r, f) < _4_ (2alog _2_ + 4log+ 1/(0)1 + b1 ) . 1-r 1-r 20 II' (0) I < 1. In this case, consider a value 0 < r < 1 and let Zo be a point of the circle Izl = r such that

(2.31)

= re iOo

I/(zo) I = M(r, f) . Consider the segment 8 : z = te iOo (0 ::::: t ::::: r) and distinguish two cases: 1) On 8 we always have 1/'(z)1 ::::: 1. Then

I/(zo) I : : : 1/(0) I + r and log M(r, f) ::::: log + 1/(0) I + log 2 .

(2.32)

2) On 8 we do not always have 1f'(z)1 ::::: 1. Then we can get a point t1eiOo (0 < t1 < r) of 8, such that on the segment 8 1 : z = te iOo (0::::: t < td we have 1f'(z)1 < 1 and II'(zdl = 1. Consider the function

Zl

=

which is holomorphic in the circle I~I < 1 and does not take the values 0 and 1. Hence by (2.30)' in the circle I~I < 1 we have

logIF(~)I< l~kl (2aIOg1~kl+410g+IF(0)1+210g+iF'~0)I+b1). (2.33) We have

59

Fix-points of Merom orph ic Functions

In (2_33)' taking in particular

Zo - ZI !:

=

(r - tt}e iOo

1 -IZII

=

1 - tl

we get 2 +4Iog+ IJ(O)I logM(r,J) < -4- ( 2alog--

1-r

1-r

+2log _1_ + bi + log 2) . 1-r

(2.34)

Consequently in the two cases, by (2.31)' (2.32) and (2.34) we have for 0< r < 1,

_2_)

log M(r, J) < _1_ (1610g+ IJ(O)I + Blog 1-r 1-r

where B is a positive numerical constant. Theorem 2.3 is therefore proved.

Theorem 2.4. Let w = J(z) be a holomorphic function in the circle 1 satisfying the condition:

Izl <

M(~,J)

J(O)=O,

2: 1.

Then in the circle Izl < 1 the function J(z) takes every value w on a certain circle Iwl = r, with r > A, A being a positive numerical constant. This is a theorem of Bohr, H. (Uber einen Satz von Edmund Landau, Scripta Univ. Hierosolymitanarum, 1 (1923)).

Proof. Consider a number R > 0 and assume that for each number r 2: R, there exists a point Wo of the circle Iwl = r, such that the function J(z) does not take the value Wo in the circle Izl < 1. Then there are two points WI and W2 respectively on the circles Iwl = Rand Iwl = 2R, such that f(z) does not take the values Wl and W2 in the circle Izl < 1. Consider the function g () Z

=

which is holomorphic in the circle and 1, and we have

f(z) W2 -

Izl

Wl Wl

< 1 and does not take the values 0

Fix.points and Factorization of Meromorphic Functions

60

Consequently by Theorem 2.3, for

Ig(z)1 < exp

Izl

< 1 we have

(1 !Izi 1:Izl) iog

and

Let Zo be a point of the circle

Izi =

1/2 such that

If(zo)1 = M

(~, f)

,

then we get

1 < kR,

k = 1 + 3e 2B log 4

•

Hence there is a circle Iwl = r, r ~ l/k, such that in the circle Izl < 1, f(z) takes every value w of the circle Iwl = r. Now we are going to prove some theorems on the growth of composite functions.

Theorem 2.5. Let g(z) and h(z) be two entire functions and let f(z) = g{h(z)}. If h(O) = 0, h(z) 'f; 0, then there is a number c (0 < c < 1) such that for r > 0, we have (2.35) This theorem is due to P6lya, G. ("On an integral function of an integral function", J. London Math. Soc., 1 (1926) 12-15).

Proof. Consider the function

which evidently satisfies the conditions of Theorem 2.4. There is then a circle Iwl = R, R ~ c (0 < c < 1), such that in the circle IZI < 1, the function H(Z) takes every value w of the circle Iwl = R. Hence in the circle Izl < r, the function h(z) takes every value W of the circle IWI = RM (~,h). Set

61

Fix-points of Meromorphic Functions

Let Wo be a point of the circle IWI = R', such that Ig(Wo)1 = M(R',g) and let Zo be a point of the circle Izl < r, such that h(zo) = Woo Then we have

M(r,f)

~

I/(zo)1 =

Ig{h(zo)} I = Ig(Wo)1

= M(R',g) ~ M {cM G,h) ,g} Corollary 2.1. Let g(z) and h(z) be two entire functions and let I(z) = g{h(z)}. IT h(z) is non-constant, there is a number CI (0 < CI < 1) such that when r is sufficiently large, we have

M(r,f) ~ M {cIM G,h) ,g} .

(2.36)

Proof. Let gdz) = g(z + h(O)), hdz) = h(z) - h(O). Then I(z) gl {hdz)} and by Theorem 2.5,

M(r,f) ~ M {cM

G,hl) ,gd

=

.

Since

g(z) = gdz - h(O)),

h(z) = hl(z) + h(O) ,

we have

M(r, g)

~

M(r + Ih(O) I, gl),

M(r, h)

~

M(r, hd + Ih(O) I .

Hence when r is sufficiently large, we have

M(r,g) and

M (r, f)

~

M(2r,gd,

M(r,h) < 2M(r, hI)

~ M { ~ M ( ~, h) ,gl } ~ M { ~ M (~ , h) , g}

Now we prove another theorem of P6lya, also based upon Theorem 2.4.

Theorem 2.6. Let I(z) and g(z) be two non-constant entire functions such that the order of the function 0 and c = 1(0). Then the function

F(Z) =

I (!rZ) - c MUr,1 - c)

Fix-points and Factorization of Meromorphic Functions

62

satisfies the condition of Theorem 2.4_ Hence there is a circle Iwl = R (R > A) such that in the circle IZI < 1 the function F(Z) takes every value w of the circle Iwl = R_ It follows that in the circle Izl < r/2 the function I(z) takes every value W of the circle IW - e I = RM (~, I - e) _ Set

el = R' such

and let Wo be a point of the circle IW -

Ig(Wo)1 and let

Zo

=

be a point of the circle

max

IW-cl=R'

Izl <

that

Ig(W)I,

r/2 such that

I(zo) = Wo . Then M

G,
max

=

IW-cl=R'

Ig{f(zo)} I = Ig(Wo)1 Ig(W)I;:::

max Ig(W)1 = M(R' IWI=R'-Icl

where r is assumed to be sufficiently large such that R'

R'

-lei>

AM

> lei.

lei, g)

We have

G,/) -lel(A+ 1) .

Since I(z) is a transcendental entire function, for any positive integer N, we have AM > r N + Ie I(A + 1)

G, I)

provided that r is sufficiently large. Hence

(2.37) By hypothesis, the order of 0 is a constant. So we have, for sufficiently large t, ..L

10gM(t,g)
•

63

Fix-points of Meromorphic Functions

Since N may be taken arbitrarily large, the order of g( z) is zero. Theorem 2.7. Let J(z) and g(z) be two transcendental entire functions and let ip(z) = g{J(z)}. Then lim T(r, ip) r--+oo

T(r, g)

= 00

.

(2.38)

Proof. In the proof of Theorem 2.6, we have obtained (2.37)' where N is any positive integer, provided that r is sufficiently large. We have

hence Since T(r, g) is a convex function of log r we see that the function

T(r, g) - T(ro, g) log r - log ro is non-decreasing for r > ro, where ro > 1. Hence

T(r N , g) - T(ro, g) > T(r, g) - T(ro, g) N log r - log ro

-

log r - log ro

and, when r is sufficiently large, we have

N

T(r N , g) ~ 2{T(r, g) - T(ro, g)} > and

N

"4 T (r, g)

,

N

T(r, ip) > 12 T(r, g) . Since N is arbitrary, we have (2.38). Theorem 2.B. Let J(z) be a transcendental entire function and g(z) a transcendental meromorphic function. Let ip(z) = g{J(z)}. Then lim T(r, ip) r--+oo

T(r, J)

= 00

.

(2.39)

64

Fix-points and Factorization of Merom orphic Functions

Proof. Assume first that g(z) has an infinite number of zeros. Choose p zeros ai(i = 1,2, ... ,p) of g(z), such that

lai - ajl > 1

(i,j = 1,2, ... ,pji"l j) .

:5

Then it is easily seen that we can find a number 0 < 8 1/2 and a number M > 0 such that in the p circles Ci : Iz - ail < 8(i = 1,2, _.. , p) we have respectively the p inequalities

Ig(z)1 < Mlz-ail

(i= 1,2,._. ,p).

Evidently any point z belongs at most to one of the circles Ci (i 1,2, ... ,pl. Consequently p

log

+1",

Ig(z)1 ~

8 log+ Iz _8 ail -log + 8M, p

log and m

+

1

1~(z)1 ~

8 log + I'(z)8_ ail-log + 8M, '"

(r,;;) ~ ~ m (r, ,~aJ -log+ 8M .

Then making use of the inequality m

(r, -, 1 ):5 m (r, _, 8 ) + log! , -at -at 8

we get (2.40) where h is a constant. On the other hand, we see easily that (2.41) Inequalities (2.40) and (2.41) yield

65

Fix-points of Meromorphic Functions

Finally making use of the relations

T (r, f

~

aJ

= T(r, f)

+ 0(1), T (r, ~)

= T(r, IP)

+A,

we get

T(r, IP) ;::: pT(r, f)

+ 0(1)

,

and when r is sufficiently large,

T(r, IP) T(r,f) >

p

2.

Since p is arbitrary, we have (2.39). In the above we have assumed that g(z) has an infinite number of zeros. In general, since g(z) is a transcendental meromorphic function, there is a finite value Wo such that the function g1(Z) = g(z) - Wo has an infinite number of zeros. Set

IPdz)

= g1 {f(z)} = g{f(z)} -

Wo

= IP(z) -

Wo .

By the result just obtained, we have

lim T(r,IPd = r-oo

00 •

T(r, f)

Then by the inequality

T(r,IPd ~ T(r, IP)

+ log+ Iwol + log 2 ,

we get again (2.39). Note that in the proof of Theorem 2.8, the condition that f(z) is a transcendental entire function is not necessary. What is important is that f(z) is a non-constant entire function. 2.3. SOME THEOREMS OF ROSENBLOOM ON FIX-POINTS For reference, see "The fix points of entire functions", Medd. Lunds Univ. Mat. Sern., Suppl.-Bd. M. Riesz 186 (1952). Definition 2.1. Let f(z) be a meromorphic function. A point zo(zo too) is said to be a fix-point of f(z), if f(zo) = zoo This is equivalent to say that Zo is a zero of the function f (z) - z. Theorem 2.9. Let P(z) be a polynomial of degree n ;::: 2 and let f(z) be a transcendental entire function. Then the function P{f(z)} has an infinite number of fix-points. Proof. Suppose, on the contrary, that the function P{f(z)} has at most a finite number of fix points. Then the function P{f(z)} - z has at

Fix-points and Factorization of Meromorphic Functions

66

most a finite number of zeros, hence

P{f(z)} - z = Q(z)ea(z)

(2.42)

where Q(z) t 0 is a polynomial and o:(z) is an entire function_ Since the left hand member of (2.42) is a transcendental entire function, so o:(z) is non-constant. The equation f{P(z)} = z has necessarily at most a finite number of solutions. In fact, if Zo is a solution of this equation, then

f{P(zo)} = Zo , hence

P(f{P(zo)}) = P(zo) and by (2.42),

Q{P(zo)} = 0 . So Zo is a zero of the polynomail Q{P(z)}. Since Q{P(z)} most a finite number of zeros. By Theorem 2.8, lim T(r, f(P)) r-+oo

hence

T(r, P)

= 00

t

0, it has at

,

f {P(z)} is a transcendental entire function and we have f{P(z)} = L(z)eP(z)

+z

(2.43)

where L(z) to is a polynomial and ,B(z) is a non-constant entire function. From (2.42) and (2.43) we get

P{L(z)eP(z)

+ z} = P(f{P(z)}) = Q{P(z)}ea{P(z)} + P(z) .

(2.44)

Let

P( Z)

= CoZ n + C1Zn-l + ... + Cn (co i' 0,

n? 2) .

After calculation, we get n

L i=l

ui(z)eiP(z)

+ v(z)ea{P(z)}

= 0

(2.45)

Fix-points of Meromorphic Functions

67

where uj(z)(i = 1,2, ... , n) and v(z) = -Q{P(z)} ~ 0 are polynomials. It is easy to see that the polynomial Uj(z) and the polynomial (2.46) have the same degree and have the same coefficient for the term of the highest degree. Consequently the polynomials Uj(z) U = 1,2, ... , n) and v(z) are all non-identically equal to zero. N ow distinguish two cases: 1) The functions J'{3(z) - a{P(z)}(i = 1,2, ... , n) are all non-constant. In this case, by (2.45) and Theorem 1.7, uj(z)(i = 1,2, ... , n) and v(z) are all identically equal to zero. So we get a contradiction. 2) There is an integer io (1 ~ io ~ n) such that io/3(z) - a{P(z)} is a constant. In this case, for i =I- J'o (1 ~ i ~ n)'J'/3(z) - a{P(z)} is nonconstant, for otherwise, /3(z) would be a constant. Writing (2.45) in the form n

L

I

uj(z)ej.B(z) + vdz)eo{P(z)} = 0

j=l

2:'

the term ui(z)eio.B(z) is omitted. Since n ~ 2 and i/3(z) - a{P(z)}(1 ~ J- ~ n,i =I- J'o) are non-constant, we get again a contradiction by Theorem 1.7. where in the sum

Lemma 2.4. Let
68

Fix·points and Factorization of Meromorphic Functions

We first prove Lemma 2.5 as follows. Assume on the contrary, that there exist fi(Z), gi(z)(i = 1,2,3) satisfying the conditions 1), 2), 3). Then the identities cp(z) = h{gi(zH (i = 1,2,3) imply that gi(z)(i 2.8, we have

=

1,2,3) are non-constant. Consequently by Theorem

= o{T(r, cpH

T(r, gil

(i

= 1,2,3)

(2.47)

.

Then by Theorem 2.1, for r> 1, we have

T(r,cp)

~

t

(r,~) + o{T(r, cpH + S(r) cp g.

N

i=1

(2.48)

where S(r) satisfies the conditions 1° and 2° in Theorem 1.4 with respect to cp(z). Since by hypothesis h(z) has at most a finite number of fix-points Zik(k = 1,2, ... ,Pi), we have _N ( r, -1 -)

cp - gi

= _N ( r,

1

h(gi) - gi

) <

-

~N Pi _

(

r,

1)

gi - Zik

From (2.47), we have

N (r,

1

gi - Zik

= o{T(r,cpH (k = 1,2, ... ,Pi) ,

)

hence

N (r, _1_) = o{T(r, cpH

cp - gi

(i = 1,2,3) .

Then by (2.48) we get

T(r, cp)

~

o{T(r, cpH + S(r) .

(2.49)

Since gi(Z) is non-constant, (2.47) implies lim T(r, cp) logr

r-+oo

= 00

•

(2.50)

Next since S(r) satisfies the conditions 1° and 2° in Theorem 2.1 with respect to cp(z), we get a contradiction from (2.49) and (2.50).

69

Fix·points of Meromorphic Functions

By the same method we can prove Lemma 2.4 base upon Theorem 2.2.

Theorem 2.10. Let F(z) and G(z) be two transcendental entire functions. Then at least one of the two functions F(z) and F{G(z)} has an infinite number of fix-points. Proof. Assume, on the contrary, that the conclusion of this theorem is untrue. Consider the functions

cp(z) = F{G(z)} , fdz) = F(z), h(z) = F{G(z)}, gdz)

= G(z),

g2(Z)

=Z.

By Theorem 2.8, we have

.

T(r,cp)

r!:~ T(r, G) = 00

,

hence cp(z) is a transcendental entire function. We have

So by the assumption, the conditions 1), 2), 3) in Lemma 2.4 are all satisfied, but this is impossible.

Corollary 2.2. Let f(z) be a transcendental entire function. Then the function f{f(z)} has an infinite number of fix-points. Proof. By Theorem 2.10, at least one of the two functions f(z) and f{f(z)} has an infinite number of fix-points. Noting that if Zo is a fix-point of f(z), then f(zo) = Zo, f{f(zo)} = f(zo) = Zo, and hence Zo is also a fix-point of f{f(z)}. It follows that f{f(z)} has always an infinite number of fix-points.

Theorem 2.11. Let F(z) be a transcendental meromorphic function and G(z) and H(z) two transcendental entire functions. Then at least one of the three functions F(z),F{G(z)} and F{G[H(z)]} has an infinite number of fix-points. Assume, on the contrary, that the conclusion of this theorem is untrue. Consider the functions

cp(z)

= F{G[H(z)]}

,

70

Fix·points and Factorization of Meromorphic Functions

JI{z) = F{z)' gJ(z) = G[H{z)],

h{z) = F{G{z)}, g2{Z) = H{z)'

h{z) = F{G[H{z)]} ; g3{Z) = Z .

By Theorem 2.8, we have

T{r,gd . I1m

r-+oo

-

T{r, g2) -

00

,

hence the functions gi{z)(i = 1,2,3) are distinct and gJ(z) is a transcendental entire function. Also by Theorem 2.8, we see that h{z) and h{z) are transcendental meromorphic functions. On the other hand, evidently the identities


=

Jdgi{Z)}

(i = 1,2,3)

hold. Hence J;(z)' g;(z) (i = 1,2,3) satisfy the conditions 1), 2), 3) Lemma 2.5 and we get a contradiction.

III

2.4. SOME THEOREMS OF BAKER ON FIX-POINTS For refernce, see "The existence of fix-points of entire functions", Math. Zeit 73 (1960).

Definition 2.2. Let J{z) be an entire function. Set

JI{z) = J{z),

h{z) = JUJ(z)},

h{z) = J{h{z)}, ... ,

in general

In+J(z) = JUn{z)}

(n = 1,2, ... ) .

Let n ~ 2 be an integer. Then a fix-point of In{Z) is called a fix-point of order n of J{z); if Zo is a fix-point of order n of J{z)' but of no lower order, then Zo is called a fix-point of exact order n of J{z). A fix-point of order 1 of J{z) is also called a fix-point of exact order 1 of J{z).

Theorem 2.12. Let J{z) be a transcendental entire function. Then for each positive integer n, J{z) has an infinite number of fix-points of exact order n, except at most for one exceptional positive integer n. Proof. Note first that if n integers, then

=

h

+ k,

where hand k are two positive

In{Z) = Jd!h{z)}

Fix-points of Meromorphic Functions

71

and, by Theorem 2.8, we have

r T(r, fn) r~~ T(r, ih)

= 00

.

Now suppose that there is a positive integer n such that f(z) has at most a finite number of fix-points Zj(j = 1,2, ... ,p) of exact order n. Consider a positive integer m > n. If Zo is a root of the equation

then and hence Z = fm-n(zo) is a fix-point of fn(z). It follows that, either for a certain j(1 ~ j ~ p) we have

or Z is a fix-point of exact order no(1 ~ no ~ n - 1) of f(z), in the latter case, we have

fno(Z)

=Z ,

that is

fm-n+no (zo) = fm-n(zo) . Summing up these results, we conclude that Zo is a zero of one of the following functions:

fm-n(z) -

Zj

(j = 1,2, ... ,p) ,

fm-n+no (z) - fm-n(z)

(no

=

1,2, ... ,n - 1)

and hence

N

(r, fm - 1fm-n )

t

~ j=1 N + = 0

(r, fm-n1 -

) Zj

~-( 1 - - -) ~ N r, - - - -

no=1

fm-n+no - fm-n

{~ T(r, f.) } = o{T(r, fm)} .

(2.51)

72

Fix-points and Factorization of Merom orphic Functions

Now applying Theorem 2.2, with Im(z) for I and Im-n(z), Z for If'i(i = 1,2), we get, for r > 1,

T(r, 1m) :5 N (r, 1m _l/m_n) + N (r, 1m 1_

Z) + o{T(r, 1m)} + S(r)

where S(r) satisfies the conditions 10 and 2 0 in Theorem 2.1 with respect to Im(z). Consequently when r is exterior to a sequence of intervals of finite total length, we have

T(r, 1m) :5 N (r, 1m 1_

Z) + o{T(r, 1m)} .

(2.52)

Now divide the zeros of Im(z) - Z into two kinds: 1) A zero Zo of 1m (z) - Z belongs to the first kind, if Zo is not a zero of one of the functions f,,,(z) - z (k = 1,2, ... ,m - 1), in other words, Zo is a fix-point of exact order m of I(z). 2) A zero Zo of Im(z) - z belongs to the second kind, if Zo is a zero of one of the functions Ik (z) - z (k = 1,2, ... ,m - 1). By (2.52) and the relation

~ N (r, Ik ~ z) = 0 {~T(r'lk)} = o{T(r, 1m)} , evidently Im(z) - z has an infinite number of zeros of the first kind, hence I(z) has an infinite number of fix-points of exact order m. So far we have proved that if there is a positive integer n such that I(z) has at most a finite number of fix-points of exact order n, then for any positive integer m > n,/(z) has an infinite number of fix-points of exact order m. The conclusion of Theorem 2.12 is therefore valid.

Theorem 2.13. Let I(z) be a polynomial of degree d ~ 2. Then for each positive integer n,/(z) has at least one fix-point of exact order n, except at most one positive integer n. Proof. Evidently the degree of In (z) is d n and hence for each positive integer n,/(z) has fix-point of order n. In particular, I(z) has fix-point of order 1, and hence fix-point of exact order 1. Suppose, on the contrary, there exist two positive integers nand k such that n > k ~ 2 and that I(z) has neither fix-point of exact order n, nor fix-point of exact order k. Consider the rational function

( ) = In(z) - z In-k(Z) - z

If' z

Fix-points of Meromorphic Functions

73

Denoting by No the number of the zeros of !p(z), each zero being counted only once, we are going to prove that we have (2_53) In fact, if Zo is a zero of !p(z), then Zo is a zero of fn(z) - z_ Since by assumption, f(z) has no fix-point of exact order n, Zo is a fix-point of exact order ;(1 :5 ; < n) of f(z). n must be divisible by;. In fact, this is clear, if; = 1. IT 1 <; < nand n is not divisible by;, then m; < n < (m + 1); where m is a positive integer, hence n = m; + h (1:5 h < ;) and

Then from

h;(zo) = /;{/;(zo)} = /;(zo) = Zo , f3;(ZO) = /;{h;(zo)} = /;(zo) = Zo , fm;(zo) = Zo , we have

and get a contradiction. Consequently n is divisible by ;. Distinguish two cases: 1) n = 3. In this case, ; = 1, and the zeros of !p(z) are fix-points of f(z), whose number is at most equal to d, hence (2.53) holds. 2) n = 4. In this case, ; = 1 or ; = 2. Since a fix-point of order 1 of f(z) is also a fix-point of order 2 of f(z), it follows that the zeros of !p(z) are fix-points of h(z), whose number is at most cP, it follows that (2.53) also holds. 3) n > 4. In this case, first note that n is not divisible by n - 1 and n - 2. In fact, if n is divisible by n - 1, then n = m(n - 1) (m is a positive integer with 1 < m < n), and therefore (m - l)n = m which is obviously impossible. Similary, ifnis divisible by n-2, then n = m(n-2)(1 < m < n) and therefore (m - l)n = 2m, 2n = 2m(n - 2) = (m - l)n(n - 2),2 = (m - l)(n - 2) which is impossible. Hence;:5 n - 3 and each zero of !p(z)

74

Fix-points and Factorization of Meromorphic Functions

must be a fix-point of one of the polynomials fj(z)(j = 1,2, ... ,n - 3). Consequently

n-3 N o_~ < " dj j=1

=

dn-2 _ d < dn - 2 - d d-1-

< dn - 2 .

(2.53) therefore always holds. Now denoting by N 1 the number of zeros of cp(z) - 1, each zero being counted only once, we are going to prove that (2.54) In fact, if Zo is a zero of cp(z) - 1, then

which shows that Z = fn-k(ZO) is a fix-point of ik(z). Since, by assumption, f(z) has no fix-point of exact order k, hence Z is a fix-point of exact order j with 1 :=; j < k. As in the above, we see that k is divisible by j. Distinguish two cases: 1) k ~ 3. In this case, we see, as in the above, that k is not divisible by k - 1, hence 1 :=; j :=; k - 2. Since

fj(Z) = Z,

Z = fn-dzo)

shows that Zo is a zero of the polynomial fn-k+j(Z) - fn-k(Z), we have

k-2 N1 :=;

L

j=1

n-2 dn-k+j <

L

d j < dn - 1 .

j=1

Hence (2.54) holds. 2) k = 2. In this case, j = 1, and hence Zo is a zero of the polynomial fn-t{z) - fn-2(Z). Consequently N1 :=; dn - 1 and (2.54) also holds. Now denoting respectively by No, N1 and N' the number of zeros of cp(z), cp(z) - 1 and cp'(z), each zero is counted according to its order of multiplicity. Then we have

Fix-points of Merom orph ic Functions

75

this is because a zero of order m> 1 of cp(z) or cp(z) - 1, is a zero of order m - 1 of cpl(Z). Hence, by (2.53) and (2.54), (2.55) On the other hand, cp(z) has a pole of order dn - dn - k at infinity. Let p be the number of the other poles of cp(z), each pole is counted according to its order of multiplicity. Then the number P of the poles of cp(z) (with due count of order of multiplicity) in the extended complex plane is given by (2.56) Evidently the number of the poles of cp(z) - 1 in the extended complex plane is also P. For cpl(Z), it has a pole of order d n - dn - k - 1 at infinity and the number of its other poles (with due count of order of multiplicity) does not exceed 2p, hence in the extended complex plane, the number pI of the poles of cp' (z) satisfies the inequality pI ~ d n

_

dn -

k -

1 + 2p .

(2.57)

Now it is well known that in the extended complex plane, the number of the zeros and the number of the poles of a rational function are equal (with due count of order of multiplicity), hence from (2.56) we have

On the other hand, from (2.57) we have N'

=

pI ~ d n

-

dn -

1 + 2p .

k -

Finally from (2.55) we get 2(dn

_

dn -

k

+ p)

~

dn _ dn ~

dn - 2

+

dn - k

+

+ dn - 1 + N ' dn - 2 + d n - 1 + dn _

~ dn - 2

dn - k ~ dn - 1 _

dn - 2

1

~ 2d n - 2

~ dn - 1

and arrive at a contradiction.

+

dn - 1 -

+

+

k -

1 + 2p ,

1 -

1 ~ dn

1,

dn - 1 -

dn - 1 _

dn -

1

1 = 2d n -

-

1 ,

76

Fix-points and Factorization of Merom orphic Functions

Consider for example the entire function I{z) = e% + z which has no fix-point, hence the positive integer n = 1, is exceptional in the sense of Theorem 2.12. On the other hand, consider the polynomial f(z) = z2 - z. We have h{z) - z = z3{z - 2), hence h{z) has two fix-points z = 0 and z = 2. Since these two points are also fix-points of I{z), hence the positive integer n = 2 is exceptional in the sense of Theorem 2.13. An interesting question is that, in the case of transcendental entire functions, whether the exceptional positive integer n in Theorem 2.12 may be different from 1. 2.5. NORMAL FAMILIES OF HOLOMORPHIC FUNCTIONS In the theory of normal families of holomorphic functions, the following theorem of Montel is fundamental.

Theorem 2.14. Let In {z)(n = 1,2, ... ) be a sequence of holomorphic functions in a domain D. IT this sequence of functions is locally uniformly bounded in D, then we can extract from this sequence of functions a subsequence In" (z)(k = 1,2, ... ) which converges locally uniformly in D. To say that the sequence In{z)(n = 1,2, ... ) is locally uniformly bounded in D means that for each point Zo in D, there is a circle c : Iz - Zo I < r belonging to D and a positive number M such that

I/n{z)l $

for n ~ 1, z E c .

M

(2.58)

Similarly we define the notion of local uniform convergence in D. For the proof of this theorem, we need two lemmas.

Lemma 2.6. Let In{z)(n = 1,2, ... ) be a sequence of functions holomorphic in a circle Izl $ R. IT the sequence In{z)(n = 1,2, ... ) is uniformly bounded for Izl $ R and converges at each point of a set 8 for which the point z = 0 is a point of accumulation, then the sequence In {z)(n = 1,2, ... ) converges uniformly in any circle Izl $ r (O < r < R). To say that a function is holomorphic in a circle Izl $ R means that it is holomorphic in a circle Izl < p with p> R. Proof. Let

00

In{z)

=

Lain) zk . k=O

Since the sequence In{z)(n = 1,2, ... ) is uniformly bounded for we have

I/n{z) -

In{O) I $

I/n{z)1 + I/n{O)1 $

2M for n ~ 1,

Izl

~

Izl R

$ R,

Fix-points of Meromorphic Functions

77

where M > 0 is a constant independent of nand z. Hence by Schwarz lemma, we have

Consider a point Zo of the circle

JzJ :5 R.

We have

Given arbitarily a number e > 0, let Zo be a point of the set

8

such that

and next let N be a positive integer such that

Jfn(zo) - fm(zo)J <

e

2

(n ~ N, m ~ N) .

Then we have

Jfn(O) - fm(O)J < e

(n ~ N, m ~ N)

which shows that the sequence a~n) = fn(O)(n = 1,2, ... ) converges to a limit ao. Consider the sequence

Since on the circle

JzJ

=

R we have

this inequality also holds in the circle JzJ < R, by the maximum modulus principle. Besides this sequence also converges at each point of the set 8. Consequently as in the above, we see that the sequence a~n) = gn(O) (n = 1,2, ... ) converges to a limit al. Continuing in this way, we see that for any k, the sequence ain) (n = 1,2, ... ) converges to a limit ak.

78

Fix-points and Factorization of Meromorphic Functions

By Cauchy inequality,

Iak(nll

M

~

(2.59)

Rk '

hence (2.60) and the series 00

f(z) =

L

ak zk

k=O converges absolutely in the circle f(z). Set

Izl <

R and defines a holomorphic function

K

fn(z)

=

L

K

aLn l zk + Pn(z) ,

f(z)

=

k=O

L

ak zk + p(z)

k=O

where 00

Pn(z)

=

00

L

p(z)

L

=

k=K+l

k=K+l

Consider a circle Izl ~ r (0 < r < R). By (2.59) and (2.60), we see that in this circle the inequalities

hold, where () = r/ R. Given arbitrarily a number of e > 0, let K be a positive integer such that ()K+l e M--<1- () 3 and then let N be a positive integer such that for n K

IL k=O For n

~

K

aLn l z k -

L

akzkl <

k=O

i (Izl

N, we have

(Izl ~ r) .

N, evidently

Ifn(z) - f(z)1 < e

~

~

r)

79

Fix-points of Meromorphic Functions

and hence in the circle uniformly to f(z).

Izl

$ r, the sequence

fn(z)(n

= 1,2, ... ) converges

Lemma 2.1. Let fn(z)(n = 1,2, ... ) be a sequence of holomorphic functions in a domain D. If this sequence is locally uniformly bounded in D and converges at each point of a set s which has a point of accumulation in D, then this sequence converges locally uniformly in D. This is a theorem of Vitali.

Proof. Let Zo be a point of accumulation of the set s in D. Consider any point Z of D such that Z ¥ zoo Join Zo and Z by a polygonal line L in D. By Heine-Borel theorem, we see easily that, under the conditions of Lemma 2.7, we can find a number d> 0 having the following property: For any point ~ of L, the circle Iz - ~I $ d is interior to D and the sequence fn(z)(n = 1,2, ... ) is uniformly bounded in this circle. Now in the sense from Zo to Z, take on L a finite number of points ~i(j = 0, 1,2, ... , n) such that ki-~i-11
(j=1,2, ... ,n)'

~o=Zo,

~n=Z.

Consider first the point ~o and the circle I'o : Iz-~ol $ d. Since the sequence fn(z)(n = 1,2, ... ) is uniformly bounded in ro and converges at each point of the set s for which ~o is a point of accumulation, hence by Lemma 2.6, the sequence fn(z)(n = 1,2, ... ) converges uniformly in the circle "Yo : Iz-~ol $ d/2. Next consider the point ~1 and the circle I'1 : Iz - ~11 $ d. Since the point ~1 is interior to the circle /0 : Iz - ~o I < d/2, we see again, by Lemma 2.6, that the sequence fn(z)(n = 1,2, ... ) converges uniformly in the circle "Y1 : Iz - ~11 $ d/2. Continuing in this way, we see finally that the sequence fn(z)(n = 1,2, ... ) converges uniformly in the circle "Yn: Iz- ~nl $ d/2. Thus we have shown that the sequence fn(z)(n = 1,2, ... ) converges locally uniformly in D. N ow let us return to the proof of Theorem 2.14. First let Zk (k = 1,2, ... ) be a sequence of points in D, which has a point of accumulation in D. Since the sequence fn(zd(n = 1,2, ... ) is bounded, we can extract from it a convergent subsequence

In other words, the sequence 81

:

fal

(z), fa, (z), ...

80

Fix-points and Factorization of Meromorphic Functions

converges at the point Z1. In the same way we can extract from the sequence 8 1 a subsequence

82

:

Ifh (z), Ip2 (z),

...

which converges at the point Z2. Then from the sequence 8 2 we extract a subsequence 83 : 1'"11 (z), 1'"12 (z), . .. which converges at the point Z3. In general, from the sequence 8 k - 1 we extract a subsequence 8 k = 1>.1 (z), />'2 (z), ... which converges at the point Zk. Continuing this process we get a sequence of sequences 8 k (k = 1,2, ... ). Consider the sequence

The first term, second term and the third term of 8' are respectively the first term of 8 1 , the second term of 8 2 and the third term of 83j in general the kth term of 8' is the kth term of 8 k • Obviously for each k, beginning from the kth term of 8' all the subsequent terms of 8' belong to 8 k • Consequently the sequence 8' converges at each of the points zk(k = 1,2, ... ). On the other hand the sequence 8' is locally uniformly bounded in D, by the conditions of Theorem 2.14. Hence by Lemma 2.7, the sequence 8' converges locally uniformly in D.

Definition 2.3. Let {J(z)} be a family of holomorphic functions in a domain D. If from every sequence In(z)(n = 1,2, ... ) of this family we can extract a subsequence Ink (z)(k = 1,2, ... ) satisfying one of the following two conditions: 1) Ink (z)(k = 1,2, ... ) is locally uniformly convergent in Dj 2) As k -+ 00, Ink (z) tends locally uniformly to infinity in Dj then we say that the family {J(z)} is normal in D. In this definition, the word "family" means "set". We say that the family {J(z)} is normal at a point Zo of D, if there is a circle c : Iz - zol < r interior to D such that the family {J(z)} is normal in c. Evidently if the family {J(z)} is normal in D, then it is normal at each point of D. Conversely we have the following theorem: Theorem 2.15. Let {J(z)} be a family of holomorphic functions in a domain D. If this family is normal at each point of D, then it is normal in D.

81

Fix·points of Meromorphic Functions

For the proof of this theorem we need the following lemma:

Lemma 2.8. Let {J(z)} be a family of holomorphic functions in a domain D and assume that {J(z)} is normal at each point of D. Let 8: fn(z)(n = 1,2, ... ) be a sequence of the family {J(z)} and Zo and zb two points of D. IT the sequence 8 is uniformly convergent in a circle C with center Zo, then there is a subsequence 8' of 8, such that 8' is uniformly convergent in a circle c' with center z~. Proof. Join the two points Zo and zb by a polygonal line L in D. By the conditions of Lemma 2.8 and Heine-Borel theorem, it is easy to see that we can find a number d > 0 such that for any point ~ of L, the circle Iz - ~I < d is interior to D and in this circle the family {J(z)} is normal. Now in the sense from Zo to zb take on L a finite number of points ~i(j = 0,1,2, ... , n) such that

ki -

~i-11 < d

(j = 1,2, ... ,n)'

~o

=

ZO,

~n

=

z~.

Consider first the circle Co : Iz - ~o I < d and the sequence 8. Since the family {f(z)} is normal in Co, we can, by the conditions of Lemma 2.8, extract from 8 a subsequence 8 0 which is locally uniformly convergent in Co. Next consider the circle C1 : Iz - ~11 < d and the sequence So. Since the point ~1 is interior to Co, we see that we can extract from 8 0 a subsequence 8 1 which is locally uniformly convergent in C1. Continuing in this way we get finally a sequence Sn which is locally uniformly convergent in the circle Cn : Iz - ~n I < d. Evidently this sequence Sn has the required properties of the sequence 8' in Lemma 2.8. Now let us return to the proof of Theorem 2.15. First of all, we can find a sequence of points Zi (j = 1,2, ... ) of D such that each point of D is a limiting point of this sequence. Such a sequence of points of D can be obtained in different ways of which the simplest is to take the set of rational points (points of the form z = r1 + ir2, r1, r2 being rational numbers) of D. We know that this set is countable. Consider a point zl' By hypothesis there is a circle c : Iz - zi I < r interior to D such that the family {J(z)} is normal in c. Let Ri be the least upper bound of the set of such positive numbers r. According to Ri is finite or infinite, denote by c} respectively the circle Iz - Zl I < R} /2 or the circle Iz - zil < 1. The circle c} so defined is interior to D and the family {J(z)} is normal in Ci'

82

Fix-points and Factorization of Meromorphic Functions

Now let 8: In{z)(n

=

1,2, ... ) be a sequence ofthe family {J{z)}. From

8 we can extract a subsequence 8 1 which satisfies one of the two conditions in Definition 2.3 in C1. Then from 8 1 we can extract a subsequence 8 2 which satisfies one of the two conditions in Definition 2.3 in C2. Continuing in this way and finally taking the diagonal sequence, as in the proof of Theorem 2.14, we get a subsequence 8' of the sequence 8 such that in each circle cj,8' satisfies one of the two conditions in Definition 2.3. We are going to show that the sequence 8' satisfies in D one of the two conditions in Definition 2.3. Distinguish two cases: 10 8' is locally uniformly convergent in C1. In this case, by Lemma 2.8, 8' is locally uniformly convergent in each circle Cj. Consider a point Zo of D. By hypothesis, the family {J{z)} is normal in a circle r : Iz - zol < p (O < P < 1) interior to D. Let Zj be such that IZj - zol < p'{p' < p/4). Then the circle Iz - Zjl < 2p' is interior to r, and hence the family {J{z)} is normal in this circle. Consequently, if R j is finite, then

,

2p ~ R j

,

'< 2 Rj '

p -

and hence the circle "t : Iz - zjl < p' is interior to the circle Cj. IT R j is infinite, then since p' < 1, "t is also interior to Cj. So the sequence 8' is locally uniformly convergent in the circle "t. Since Zo is a point in the circle "t, hence the sequence 8' is uniformly convergent in a circle with center zoo 2 0 As k -+ 00,8' converges locally uniformly to infinity in C1. In this case, by Lemma 2.8, in each circle cj,8' converges locally uniformly to infinity as k -+ 00. Then, as in the above it can be proved that for each point Zo of D, there is a circle with center Zo, in which 8' converges uniformly to infinity as k -+ 00.

Theorem 2.16. Let {J{z)} be a family of holomorphic functions in a domain D. Let a and b (a =1= b) be two finite values. IT each function I{z) of the family {I (z)} does not take the values a and b in D, then the family {J{z)} is normal in D. This is an important theorem of Montel (Leltons sur les Familles Normales de Fonctions Analytiques et leurs Applications, Paris, 1927). Proof. In view of Theorem 2.15, we need only to consider the case that D is a circle C : Iz - Zo I < R. Evidently it is sufficient to show that from any sequence In{z)(n = 1,2, ... ) of the family {J{z)} we can extract a subsequence Ink (z)(k = 1,2, ... ) satisfying one of the following two conditions:

83

Fix-points of Meromorphic Functions

1) Ink (z)(k = 1,2, ... ) is uniformly convergent in the circle c' : Iz-zo I < R/4. 2) As k -+ 00, Ink (z) tends uniformly to infinity in c'. To see this, consider the sequence In (zo)(n = 1,2, ... ) and distinguish two cases: 1° The sequence In(zo)(n = 1,2, ... ) is bounded:

I/n(zo)1 ::; M

(n = 1,2, ... ) .

Applying Theorem 2.3 to the function

F(Z)

= In(zo + RZ) b-a

- a

(IZI <

1) ,

we get log I/n(z) -

Ib- aiR al < R -Iz - zol x (AIog+ I/n(zo) -

Ib - al

In particular for

al + Blog

2R

R

-Iz - zol

) (2.61)

Iz - Zo I < R/2, we have

Hence the sequence In(z)(n = 1,2, ... ) is uniformly bounded in the circle Iz - zol < R/2. Then by Theorem 2.14 and Heine-Borel Theorem, we see that we can extract from the sequence In(z)(n = 1,2, ... ) a subsequence Ink (z)(k = 1,2, ... ) which is uniformly convergent in the circle c'. 2° The sequence In (zo)( n = 1,2, ... ) is unbounded. Then we can find a subsequence Ink (zo)(k = 1,2, ... ) such that

We are going to show that, as k -+ 00, Ink (z) tends uniformly to infinity in the circle c'. In fact, consider a point Zl of the circle c'. Then the circle Iz - zll < ~R is interior to the circle c. By the same method of proof of (2.61)' we get

84

Fix-points and Factorization of Meromorphic Functions

where

Rl =

~ R_ In particular in the circle

Since Zo is a point of the circle

Iz - zll < R/4, we have

Iz - zll < R/4, we have

This inequality being true for any point Zl of the circle Ink (z) tends uniformly to infinity in the circle c' as k -+

c',

it follows that

00.

Corollary 2.3. Let {J(z)} be a family of holomorphic functions in a domain D. If each function I(z) of the family {J(z)} does not take two finite values aU) and bU) such that

laU)1

IbU)1

< M,

< M,

laU) - bU)1 > 8

(2.62)

where M > 0,8 > 0 are two constants independent of I(z), then the family {/(z)} is normal in domain D. Proof. Let In(z)(n

=

1,2, ... ) be a sequence of the family {J(z)}. Set

an = a(ln),

bn = b(ln)

(n= 1,2, ... )

and consider the sequence

(n

=

1,2, ... ) .

(2.63)

gn(z) does not take the values 0 and 1 in D, hence by Theorem 2.16, the family {gn(z)(n = 1,2, ... )} is normal in D, consequently we can get a subsequence gnk (z)(k = 1,2, ... ) satisfying one of the two conditions in Definition 2.3. If gnk(z)(k = 1,2, ... ) satisfies the first of these two conditions, then from (2.62) and (2.63)' the sequence Ink (z)(k = 1,2, ... ) is locally uniformly bounded in D and then by Theorem 2.14, we can extract from the sequence Ink (z)(k = 1,2, ... ) a subsequence Im, (z)(l = 1,2, ... ) which is locally uniformly convergent in D. If gnk (z)(k = 1,2, ... ) satisfies the second condition in Definition 2.3, then evidently Ink (z)(k = 1,2, ... ) also satisfies the same condition.

Fix-points of Meromorphic Functions

85

We shall need also the following lemma:

Lemma 2.9. Let {f(z)} be a normal family of holomorphic functions in a domain D. Let fn(z)(n = 1,2, ... ) be a sequence of this family. IT for a point Zo of D, the sequence fn(zo)(n = 1,2, ... ) is bounded, then the sequence fn(z)(n = 1,2, ... ) is locally uniformly bounded in D. Proof. Let ~o be a point of D, and let c : Iz - ~ol ~ r be a circle interior to D. It is sufficient to show that the sequence fn(z)(n = 1,2, ... ) is uniformly bounded in c. In fact, if this is untrue, then to each positive integer P corresponds a positive integer mp such that

From the sequence gp(z) = fmp (z)(p = 1,2, ... ) we can extract a subsequence gpo (z)(s = 1,2, ... ) satisfying one of the two conditions in Definition 2.3. Since the sequence gpo (zo)(s = 1,2, ... ) is bounded, the sequence gpo (z) (s = 1,2, ... ) must satisfy the first condition in Definition 2.3. But this is incompatible with the fact maxzEc Igp.(z)1 > P,. 2.6. FATOU'S THEORY ON THE FIX-POINTS OF ENTIRE FUNCTIONS

Let f(z) be a transcendental entire function and let the sequence fn(z)

(n

= 1,2, ... ) be defined as follows:

h(z) = f(z), fn(z) = f{fn-l(Z)}

(n= 2,3, ... ).

Consider a point Zo at which the family {fn(z)(n = 1,2, ... )} is not normal, in other words, there does not exist a circle Iz - Zo I < r in which the family {fn(z)(n = 1,2, ... )} is normal, then Zo is called a Julia point of the family {fn(z)(n = 1,2, ... )}. The set of all Julia points of the family {fn(z)(n = 1,2, ... )} is called the Julia set of the function f(z) and is denoted by JU). The set JU) plays an important role in Fatou's theory. It will be shown that J(f) is non-empty and has other properties. At present we first prove some simple properties of the set JU).

Theorem 2.11. IT a point Zo E JU), then f(zo) E JU). Proof. Assume that f(zo) ~ JU). Then there is a circle c : Iz- f(zo) I < R in which the family {fn(z)(n = 1,2, ... )} is normal. Let "'f: Iz - zol < r be a circle such that in 'Y we have

If(z) - f(zo)/ < R/2 .

(2.64)

86

Fix-points and Factorization of Merom orphic Functions

Consider a sequence fnk (z)(k = 1,2, ... ; nk ~ 2) of the family {fn(z)(n = 1,2, ... )}. By assumption, we can extract from the sequence fnk-dz)(k = 1,2, ... ) a subsequence fmIL-dz)(h = 1,2, ... ) which converges uniformly in the circle Iz - f{zo) I < R/2 to a holomorphic function g{z) or to the constant 00. By (2.64)' evidently in the circle " the sequence fmIL (z) = fmIL-df{z)}(h = 1,2, ... ) converges uniformly to the holomorphic function g{f{z)} or to the constant 00. Hence the family {fn(z)(n = 1,2, ... )} is normal in the circle" and we get a contradiction. Theorem 2.18.

IT a point Zo E J(J) and Zl is a point such that

f{zt} = Zo then Zl E J(J). Proof. Assume that Zl ¢. J(J). Then there is a circle Cl : Iz - zll < rl such that the family {fn(z)(n = 1,2, ... )} is normal in Cl. Let, : Iz-zol < p be a circle such that the values taken by the function f{z) in the circle Iz - zll < rl/2 cover ,. Now consider a sequence fnk{z)(k = 1,2, ... ) of the family {fn(z)(n = 1,2, ... )}. By assumption, we can extract from the sequence fnddz)(k = 1,2, ... ) a subsequence fmn+l{Z)(h = 1,2, ... ) which converges uniformly in the circle Iz - zll < rl/2 to a holomorphic function or to the constant 00. In the first case, given arbitrarily a positive number e, we can find a positive integer H such that when h ~ H, h' ~ H, the inequality

holds in the circle Iz - zll < rl/2. Since

and by the choice of the circle " we have

IfmIL (z) - fmILI {z)1 < e in ,. Hence in " the sequence fmIL (z)(h = 1,2, ... ) converges uniformly to a holomorphic function. In the second case, we see in the same way, that the sequence fmIL (z)(h = 1,2, ... ) converges uniformly to the constant 00 in ,. Consequently the family {fn(z)(n = 1,2, ... )} is normal in " and we get a contradiction. Since the set J(J) has the properties in Theorem 2.17 and 2.18, we say

87

Fix-points of Meromorphic Functions

that J(J) is completely invariant under the substitution (z, f(z)).

Theorem 2.19. J(Jrn)

= J(J) (m =

2,3, ... ).

Proof. Let m ~ 2 be an integer and set g(z) = frn(z). Then we have

and in general

gn(z)

=

fnrn(z)

(n

= 1,2, ... ) .

Let Zo E J(g) and assume Zo tF. J(J). Then there is a circle c : Iz-zo I < r in which the family {fn(z}(n = 1,2, ... )} is normal. Since the functions of the family {gn(z}(n = 1,2, ... )} belong to the family {fn(z}(n = 1,2, ... )}, hence the family {gn(z}(n = 1,2, ... )} is also normal in c, and we get a contradiction. Now let Zo E J(J) and assume Zo tF. J(g). Then there is a circle c : Iz - Zo I < r in which the family {gn(z}(n = 1,2, ... )} is normal. By Corollary 2.2, the function f2 (z) has an infinite number of fix points. Let 0: and (3 (0: i (3) be two fix-points of h(z). Since the family {fn(z}(n = 1,2, ... )} is not normal in c, by Theorem 2.16, there is a function fp(z) of this family, which takes at least one of the two values 0: ane (3 in c. Let this value be 0:, so that there is a point ~ of c, such that

Consider an integer n ~ p, then n = p + 2h or n = p + 2h + 1 where h is an integer. In the first case

~

0

and in the second case,

Hence the sequence fn(~}(n = 1,2, ... ) is bounded and a fortiori, the sequence gn(~) = fnrn(~}(n = 1,2, ... ) is also bounded. By Lemma 2.9, the sequence gn (z) (n = 1,2, ... ) is uniformly bounded in the circle c' : Iz- Zo I ~ r/2, accordingly in c' we have Ign(z)I~R

(n=I,2, ... )

(2.66)

Fix-points and Factorization of Meromorphic Functions

88

where R > 0 is a constant independent of nand z. Now let n ~ m, then Am ~ n < (A + 1)m where A integer, and we have n

=

Am + jJ,

0

~ jJ ~

>

1 is a positive

m - 1.

Next we have (2.67) From (2.66) and (2.67), we deduce that the sequence fn(z)(n = 1,2, ... ) is uniformly bounded in the circle c', this contradicts the condition Zo E J (j). Now we are going to make a classification of fix-points. In what follows, f(z), fn(z) and J(j) have the same meaning as above. IT Zo is a fix-point of the function f(z), that is to say

f(zo) = Zo , then three cases are possible. 1) 1f'(zo)1 < 1. In this case, the fix-point Zo is said to be 2) If' (zo) I > 1. In this case, the fix-point Zo is said to be 3) 1f'(zo)1 = 1. In this case, the fix-point Zo is said to be In the third case, we have f'(zo) = e21rOi , where 0 ~ () number, so we can again distinguish two cases: a) () is a rational number. b) 0 is an irrational number. For the three kinds of fix-points, we have respectively theorems.

attractive. repulsive. neutral. < 1 is a real

the following

Theorem 2.20. IT Zo is an attractive fix-point of f(z), then we can find a circle c: Iz-zol < r in which the sequence fn(z)(n = 1,2, ... ) converges uniformly to Zo. Proof. By hypothesis, we have If'(zo) I < 1. Take a number h such that

If'(zo) I < h < 1 and let r be a positive number such that for 0 <

- Zo I = If(z) If(z) z - Zo z-

Iz - zol <

I

f(zo) < h . Zo

r, we have

89

Fix·points of Merom orphic Functions

Then in the circle c : Iz - zol < r, we have

If(z) - zol :'5 hlz - zol , and a fortiori,

If(z) - zol :'5 hr < r . This shows that when z E c, we have f(z) E c. Hence in c, we have

Ih(z) - zol:'5 hlf(z) - zol :'5 h21z - zol , Ih(z) - zol :'5 h 3 1z - zol , and in general, in c we have

Hence in c the sequence fn(z)(n = 1,2, ... ) converges uniformly to

ZOo

Theorem 2.21. IT Zo is a repulsive fix-point of f(z), then Zo E JU). Proof. By hypothesis, we have 1f'(zo)1 > 1. We have f~(zo)

= !'U(zo)} !,(zo) = U'(zO)}2 ,

f~(zo) = !'{h(zo)}f~(zo) =

U'(zO)}3 ,

and in general f~(zo)

= U'(zo)}n (n =

1,2, ... ) .

(2.68)

Assume, on the contrary, there is a circle c : Iz - Zo I < r, in which the family Un (z)(n = 1,2, ... )} is normal. Since

fn(zo)

= Zo (n = 1,2, ... )

,

(2.69)

by Lemma 2.8, we can get a positive number M such that in the circle Iz - zol < r/2, we have

Ifn(z)l:'5 M

(n = 1,2, ... ) .

Then by Cauchy's inequality, we have

~lf~(zo)1 = ~1f'(zo)ln :'5

M ,

Fix-points and Factorization of Meromorphic Functions

90

which is impossible, because If'(zo)1 > 1.

Lemma 2.10. IT Zo is a fix-point of f(z) such that f'(zo) = 1, then

Zo

E

JU).

Proof. By hypothesis, we \ave

f(z) = Zo + (z - zo) + a(z - zo)m + b(z - zo)m+1 + ... where m 2: 2, a =1=

o.

(2.70)

We are going to show that in general we have

fn(z) = Zo + (z - zo) + na(z - zo)m + bn(z - zo)m+l + ...

(2.71)

In fact, if (2.71) holds for an integer n, then

fn(z) - Zo = (z - zo){l + na(z - zo)m-lt/l(z)} ,

t/I(zo) = 1 ,

and hence

fn+1(z) - Zo = fn{j(z)} - Zo = {j(z) - zo}{l + na[J(z) - zo]m-lW(z)} , W(zo)=l. On the other hand, by (2.70), we have

f(z) - Zo = (z - zo){l + a(z - zo)m-lcp(z)} ,

cp(zo) = 1 ,

hence

fn+dz) - Zo = (z - zo){l + a(z - zo)m-lcp(z)} X {I + na(z - zo)m-l[l + a(z - zo)m-lcp(z)]m-lw(z)} = (z - zo){l + (z - zo)m-lcp(z)}{l + na(z - zo)m-lcpdz)} , CPl (zo) = 1 .

Hence

fn+dz) - Zo

=

(z - zo){l + (z - zo)m-l4>(z)} ,

4>(zo)

=

(n + l)a .

So (2_71) also holds for n + 1. Finally by the method used for the proof of Theorem 2_21, we see that Zo E JU).

Theorem 2.22. IT Zo is a fix-point of f(z) such that f'(zo) where fJ(O ::; fJ < 1) is a rational number, then Zo E JU). Proof. By Lemma 2.10, we may assume 0 < fJ < 1, fJ and q are two positive integers. By (2.68), we have f~(zo)

= {j'(zo)}q = e21rpi =

1,

=

= e21r8i ,

p/q, where p

Fix -points of Meromorphic Functions

91

and by (2.69), Zo is a fix-point of Jq(z), hence by Lemma 2.10, Zo E J(fq) and then by Theorem 2.19, Zo E J(f). It remains to study neutral fix-points for the case b). For this purpose, we need the following lemma:

Lemma 2.11. Let a and b be two real numbers and let e > 0 be a number. Then there exist two integers, m and n not both equal to zero, such that (2.72) Ima+ nbl < e.

Proof. Assume, on the contrary, that such two integers m and n do not exist. Set Xm,n = ma + nb. Then we have

IXm,nl ~ e for any pair of integers (positive, negative or zero) (m, n) not both equal to zero. For two such pairs of integers (ml' nd and (m2' n2) distinct from each other, we have also (2.73) Now take an integer N > 1 and consider the set S of the pairs of integers (p, q) such that 1 5 P 5 N, 1 5 q 5 N . The total number of such pairs (p, q) is N2. corresponds the interval

1p,q .. [Xp,q - ~3' X p,q Evidently

Ip,q c [-

G+

2N

A),

To each pair (p, q) of S,

+ ~] 3 ~ + 2N A]

where A = max(lal, Ibl). On the other hand, by (2.73), corresponding to two distinct pairs (PI, ql) and (p2, q2) of S, the intervals I p1q1 and Ip"q, have no common point. Consequently the sum of the lengths of the N 2 intervals Ip,q is less than the length of the interval [- (! + 2N A) , ~ + 2N A], hence

92

Fix-points and Factorization of Meromorphic Functions

Since the right member of this inequality tends to zero, as N a contradiction.

00,

we get

Theorem 2.23. Let Zo be a fix-point of I(z). IT I'(zo) = e2rr9i where is an irrational number such that 0 < (J < 1, and if Zo ¢ J(f), then there exist a circle c : Iz - Zo I < r and a sequence of positive integers Ap(p = 1,2, ... ) such that in c, the sequence f>.p (z)(p = 1,2, ... ) converges uniformly to z. (J

Proof. Apply Lemma 2.11 to the case a = (J, b = 1 then for each positive integer k, there are two integers mle and nle not both equal to zero, such that

(2.74) In this inequality, mle cannot be equal to zero, because if mle = 0, then nle i- 0 and Inlel < 11k. We may assume mle > o. Consequently there exists a sequence of positive integers mle(k = 1,2, ... ) such that lim e2rrm ,,9i = 1 . Ie-co

(2.75)

On the other hand, by hypothesis, there is a circle c : Iz - Zo I < R in which the family {fn(z)(n = 1,2, ... )} is normal. Since In(zo) = zo(n = 1,2, ... ), we can extract from the sequence 1m" (z)(k = 1,2, ... ) a subsequence b. p (z)(p = 1,2, ... ) which converges uniformly to a holomorphic function g(z) in the circle "1 : Iz - zol < r (r = R/2). We are going to show that

g(z) = z. First of all, by (2.75) and the formula

we have

I(zo) = p-co lim Ii p (Zo) = 1 . Now assume, on the contrary, that the identity g(z) = z is not satisfied. Then in the circle "1, we have

g(z) = Zo + (z - zo) + b(z - zo)m + b'(z - zo)m+l + ... where b i- 0, m

~

(2.76)

2. On the other hand, we have

(2_77)

Fix-points of Meromorphic Functions

93

where In the formula

let p -

00,

we get

J{g(z)} = g{f(z)} . Both sides of this formula are holomorphic functions in a circle " : Iz- Zo I < r' (0 < r' < r). Let us find out the coefficients of the term (z - zo)m in the expansions of J{g(z)} and g{f(z)}. First by (2.76), we have

g(z) - Zo = (z - zo){1 + b(z - zo)m-l

(2.78)

Next by (2.77), we have

+ p.{g(z) - zo} + ... + an {g(z) - zo}n + . ..

(2.79)

{g(z) - zo}n = (z - zo)n{1 + b(z - zo)m-l
(2.80)

J {g(z)} = Zo Noting that

and for n ~ 2, we have n + m - 1 ~ m + 1, we see from (2.79) and (2.80)' that in the expansion of J{g(z)} the coefficient of the term (z - zo)m is am + p.b. On the other hand, by (2.77)' we have

f(z) - Zo = p.(z - zo)t/J(z),

t/J(zo) = 1 .

Then by (2.78), we have

g{f(z)} - Zo = {f(z) - zo} + bp.m(z - zo)mW(z),

W(zo) = 1 .

Hence in the expansion of g{f(z)} the coefficient of the term (z - zo)m is + bp.m. Consequently

am

am

+ p.b = am + bp.m , p.m-l

=

1.

94

Fix-points and Factorization of Meromorphic Functions

This is impossible, because () is an irrational number. So we arrive at a contradiction. Concerning the set J(J) the following theorem of Fatou is fundamental which shows the relationship between J(J) and the fix-points.

Theorem 2.24. The set J(J) has the following properties: 1° J (J) consists of an infinite number of points and is unbounded. 2° J(J) is a perfect set, in other words, J(J) = {J(J)}'. 3° Let En be the set of the fix-points of fn{z) and define 00

E(J) =

U En . n=l

Then each point of J(J) is a point of accumulation of E(J). For the sake of convenience, let us recall the definition of the notion of point of accumulation. Let 8 be a set of points of the complex plane. A point Zo is called a point of accumulation of 8, if in each circle Iz - zol < r there is a point z' I=- zo, z' E 8. The set of the points of accumulation of 8 is denoted by 8'. If 8 = 8', then 8 is said to be perfect (8 is assumed non-empty). For the proof of Theorem 2.24, we need the following lemma:

Lemma 2.12. Let D be a domain. Assume that there exist three distinct points aI, a2 and b of D, such that 1) al and a2 are fix-points of f{z). 2) f{b) is equal to one of al and a2. Then there is a point Zo of D such that Zo E J(J). Proof. Distinguish four cases: 1° Among the two points al and a2, there is at least one, say aI, which is a repulsive fix point of f{z). Then by Theorem 2.21, al E J(J). 2° Among the two points al and a2, there is at least one, say aI, which is a neutral fix-point of f{z) and belongs to the kind a). Then by Theorem 2.2, al E J(J). 3° al is an attractive fix-point of f{z). Then by Theorem 2.20, there is a circle c : Iz - all < r in which the sequence fn{z}(n = 1,2, ... ) converges uniformly to al. We may suppose that c c D. Now assume, on the contrary, that there does not exist a point Zo of D such that Zo E J(J). Then by Theorem 2.15, the family Un{z}(n = 1,2, ... )} is normal in D. Consequently we can get a subsequence fnk (z}(k = 1,2, ... ) which

Fix-points of Meromorphic Functions

95

converges locally uniformly to a holomorphic function g(z). Since in the circle c, we have g(z) = aI, hence g(z) = al in D. On the other hand, from In(a2) = a2 (n = 1,2, ... )' we have g(a2) = a2' So we get contradiction. 4 0 al is a neutral fix-point of J(z) and belongs to the kind b). Assume, on the contrary, that there does not exist a point Zo of D such that Zo E J(f). Then the family {fn(z)(n = 1, ... 2, ... )} is normal in D. By Theorem 2.23, we can get a circle c : Iz - all < r interior to D and a sequence of positive integers .Ap(p = 1,2, ... ), such that the sequence h p (z)(p = 1,2, ... ) converges uniformly to z in c. From the sequence hp(z)(p = 1,2, ... ) we can extract a subsequence J,..(z)(q = 1,2, ... ) which converges locally uniformly to a holomorphic function g(z) in D. Since g(z) = z in the circle c, hence g(z) = z in D. In particular, g(b) = b. On the other hand, by hypothesis, J(b) is equal to one of al and a2, say J(b) = a2, hence

h(b) = J{f(b)} = J(a2) = a2 , h (b) = J {h (b)} = J (a2) = a2 , and in general,

In(b)=a2

(n=I,2, ... ).

(2.81)

Hence g(b) = a2 and we get a contradiction. Now let us return to the proof of Theorem 2.24. First suppose that J(z) has an infinite number of fix-points. To prove the part 10 of Theorem 2.24, consider a positive number R. Since the function J(z) - z has an infinite number of zeros, let al and a2 be its two distinct zeros such that lail > R (j = 1,2,). Again since J(z) is a transcendental entire function, one at least of al and a2, say aI, is such that J(z) - al has an infinite number of zeros, hence J(z) - al has a zero b with Ibl > R, b ¥- aj (j = 1,2). Take a number R' > max(lall, la21, Ibl)' then in the domain r : R < Izl < R' there are three distinct points aI, a2, b satisfying the conditions 1) and 2) in Lemma 2.12. Consequently in r there is a point Zo E J(f), and the part lOin Theorem 2.24 is proved. To prove the part 2 0 in Theorem 2.24, we first show that J(f) c {J(f)}'. Consider a point Zo E J(f) and a circle c : Iz - Zo I < r. We are going to show that in c there is a point z' such that

z'

¥- Zo,

z' E J(f) .

(2.82)

96

Fix-points and Factorization of Meromorphic Functions

In fact, by hypothesis the family {fn(z)(n = 1,2, ... )} is not normal in c. Consequently there is a subsequence Ink (z) (k = 1,2, ... ) satisfying the following property: From the sequence Ink (z)(k = 1,2, ... ) we cannot extract a subsequence which converges locally uniformly in c to a holomorphic function or to the constant 00. Consider the sequence Ink (zo)(k = 1,2, ... ) from which we can extract a subsequence 1m" (zo)(h = 1,2, ... ) converging to a limit ,x finite or infinite. Evidently the sequence 1m" (z) (h = 1,2, ... ) has also the above property. From the sequence 1m" (z)(h = 1,2, ... ) we cannot extract a subsequence which converges locally uniformly in c to a holomorphic function or to the constant 00. Now distinguish two cases: 1) ,x is finite. Take two domains r : R < Izl < R' and r 1 : Rl < Izl < R~ such that l,xl < R < Rl and that in r there is a point ~o E J(j) and in r 1 a point ~l E J(j). This is possible by part 1° in Theorem 2.24. Let H be a positive integer such that

I/m,,(Zo)1 < R

for h ~ H .

(2.83)

By Corollary 2.3 and the above property of the sequence 1m" (z) (h = 1,2, ... ), there exists an integer h ~ H such that the values taken by the function 1m" (z) in c cover one of the domains r or r 1, say r. Consequently there is a point z' of c such that 1m" (z') = ~o. By Theorem 2.19, ~o E J(jm,,), and then by Theorem 2.18, z' E J(jm,,) = J(j). Hence z' satisfies the condition (2.82). 2) ,x is infinite. Again take two domains r : R < Izl < R' and r 1 : Rl < Izl < R~ such that R' < Rl and that in r there is a point ~o E J(j) and in rIa point ~l E J (j). Next take a positive integer H such that (2.84) and then in the same way as in the above case, we see that there is a point z' of c satisfying the condition (2.82). In the above we have shown that J(j) c {J(j)}'. Conversely it is evident that {JU)}' c J(j). The part 2° in Theorem 2.24 is proved. It remains to prove the part 3° of Theorem 2.24. Let Zo E J(j). We know already that Zo is a point of accumulation of J(j). Consider a circle c : Iz - Zo I < r. Then we can get four distinct points Q; (j = 1,2,3,4) of c, such that Q; '" Zo, Q; E J(j)

(j = 1,2,3,4) .

Fix·points of Meromorphic Functions

97

By the complement of Corollary 1.6, among the four values Ot; (j = 1,2,3,4) there are at most two which are completely multiple values of f(z). So we may assume that Otll Ot2 are not completely multiple values of f(z). Among the two values Ot1, Ot2, there is at least one, say QlI such that the function f(z) - Q1 has an infinite number of zeros. There is one such zero P whose order is 1, namely,

f(P) =

f(P) of 0 .

Otll

Consequently we can get a circle.., : g(z) in .." satisfying

<

IZ-Ot11

p and a holomorphic function

f{g(z)} = z

(2.85)

in..,. We may assume that p is sufficiently small, such that.., is interior to c and that.., does not contain the point zoo We are going to show that there exists a point Zl of .., such that Zl

E

E(f) .

(2.86)

Distinguish two cases: 1) There is a point Zl of.., such that g(zt) = Zl. Then f(zt) = Zl, and hence Zl satisfies (2.86). 2) In .." there is no point Zl such that g(Zl) = Zl. Then the functions

F.n(z) = fn(z) -

Z

(

9 (z ) - z

n = 1,2, ... )

are holomorphic in..,. Assume, on the contrary, that in .., there is no point Zl satisfying (2.86). Then by (2.85), we see that in .., the functions Fn(z)(n = 1,2, ... ) do not take the values 0 and 1. It follows that the family {Fn(z)(n = 1,2, ... )} is normal in..,. This implies that the family {fn(z)(n = 1,2, ... )} is normal in..,. But Q1 E J(f) so we get a contradiction. Thus we have shown that there exists a point Zl of c, such that Zl

of

ZO,

Zl E

E(f) .

Hence Zo is a point of accumulation of E(f). In the proof of Theorem 2.24 given above, we have assumed that the function f(z) has an infinite number of fix-points. IT this condition is not

98

Fix-points and Factorization of Meromorphic Functions

satisfied, we first apply Theorem 2.24 to the function h (z). Since J (f) = J(h), hence Theorem 2.24 still holds. Theorem 2.24 shows that the set J(f) is closely related to the fix-points of all orders of the function f(z). So there is interest to study the distribution of J(f) in the complex plane. In this respect we have the following interesting theorem.

TheoreIIl 2.25. IT the set J(f) has an interior point, in other words, there exists a circle c : Iz - Zo I < r belonging to J (f), then J (f) is the complex plane. For the proof of this theorem we need the following lemma: LeIIlIIla 2.13. Let Zo E J(f). Then for each finite value a, there exist a sequence of positive integers nk (k = 1,2, ... ) and a sequence of points ~k(k = 1,2, ... ) such that lim nk k-+oo

= 00,

lim ~k k-+oo

= Zo,

fnk(~k)

=a

(k

= 1,2, ... ) ,

(2.87)

except at most for one finite value a.

Proof. Consider a finite value a. IT for each positive integer N, there exist a positive integer n and a point ~ such that n 2': N, then this value a has the required property in Lemma 2.13. In fact, for each positive integer k, there exist a positive integer nk and a point ~k such that

Therefore if a finite value a does not have the required property in Lemma 2.13, then there exists a positive integer N such that for n 2': N, the function f n (z) does not take the value a in the circle Iz - Zo I < 1 It follows that if there are two finite values a and a' (a =/; a') both not having the required property in Lemma 2.13, then we can find a positive integer NI such that for n 2': N 1 , the function fn(z) does not take the values a and a' in the circle Iz - zol < 1/N1. By Theorem 2.16, the family {fn(z)(n 2': NIl} is normal in the circle Iz - Zo I < II N 1 • It is easy to see that the family {fn(z)(n = 1,2, ... )} is then also normal in the circle Iz-zol < liN!. This contradicts the hypothesis Zo E J (f).

IN.

Fix-points of Meromorphic Functions

99

c : Iz-

Now let us prove Theorem 2.25. Assume that there is a circle zol < r belonging to JU). Let a be a finite value satisfying the condition (2.87) in Lemma 2.13. Then we can get a positive integer k such that ~k E c. Hence ~k E JU) = JUnk). Since Jnk(~k) = a, we have a E JUnk) by Theorem 2.17, and therefore a E JU). It is then clear that every finite value a E JU), and so JU) is the complex plane. In view of Theorem 2.25, it is natural to ask whether we can find a transcendental entire function J(z) such that J(J) is the complex plane. The answer to this question is affirmative. For reference, see 1. LN. Baker, "Limit functions and sets of non-normality in iteration theory", Ann. Acad. Sci. Fenn. Ser. A.I. Math. 467 (1970). 2. M. Misiurewicz, "On iterates of eZ " , Ergodic Theory Dynamical Systems 1 (1981) 103-106. IT the set JU) is not the complex plane, then its complementary set NU) is non-empty. The set NU) is the set of points Zo at which the family {fn(z)(n = 1,2, ... )} is normal. Evidently NU) is an unbounded open set. From Theorems 2.17 and 2.18, it is easy to see that the set N U) is also completely invariant under the substitution (z, J(z)). A domain D c N(J) is called a component of NUl, if there does not exist a domain l:1 such that

D c l:1 c NUl,

D

=1=

l:1 .

For reference concerning the set J(J) and the components of the set NUl, see 1. H. Topfer, "Uber die Iteration der ganzen transzendenten Funktionen, insbesondere von sin z und cos z", Math. Ann. 117 (1941) 65-84. 2. LN. Baker, "Sets of non-normality in iteration theory", J. London Math. Soc. 40 (1965) 499-502. 3. LN. Baker, "Repulsive fix-points of entire functions", Math. Zeit. 104 (1968) 252-256. 4. LN. Baker, "The domains of normality of an entire function", Ann. Acad. Sci. Fenn. Ser. All (1975) 277-283. 5. L.S.O. Liverpool, "Value distribution and related questions in iteration theory", Factorization Theory of Meromorphic Functions, New York, 1982, pp. 55-69. 6. LN. Baker, "Wandering domains in the iteration of entire functions", Proc. London Math. Soc. 49 (1984) 563-576. 7. A.K. Kromonko and M. Yu. Ljubio, "Iterates of entire functions",

100

Fix-points and Factorization of Meromorphic Functions

Preprint of the Physico-Technical Institute of Low Temperature, UkrSSR Academy of Sciences, Kharkov, 1984, N6, pp. 1-37. 8. Chi-tai Chuang, "A simple proof of a theorem of Fatou on the iteration and fix-points of transcendental entire functions", Contemporary Mathematics 48 (1985). 9. I.N. Baker, "Iteration of entire functions: An introductory survey", Lectures on Complex Analysis, Singapore, 1988. 2.7. CASE OF POLYNOMIALS In what precedes, we have obtained some results on the fix-points of a transcendental entire function and of its iterates. Most of them are also valid for the case of polynomials. Consider a polynomial P(z) of degree d ~ 2:

Let Pn(z)(n

=

1,2, ... ) be the sequence defined as follows:

Pl(Z)

=

P(z), Pn(z)

=

P{Pn-dz)}

(n

= 2,3, ... )

where Pn(z) is a polynomial of degree dn . Define J(P) to be the set of points Zo at which the family {Pn(z)(n = 1,2, ... )} is not normal. By the same method used in Sec. 2.6, we can prove the following two theorems:

Theorem 2.26. If a point Zo E J(P), then P(zo) E J(P). Theorem 2.27. If a point Zo E J(P) and Zl is a point such that P(zd = Zo, then Zl E J(P). So the set J(P) is completely invariant under the substitution {z, P(z)}. Theorem 2.28. J(Pm)

= J(P)(m = 2,3, ... ).

Proof. Let m ~ 2 be an integer. By the same method used in the proof of Theorem 2.19, we see that if Zo E J(Pm), then Zo E J(P). Conversely let Zo E J(P). To show that Zo E J(Pm), we again use the method employed in the proof of Theorem 2.19. We see that it is sufficient to prove that at least one of the polynomials P( z) or P2 (z) has two fix-points a and 13 (a i= 13). To see this, assume, on the contrary, that each of P( z) and P2 (z) has only one fix-point (finite). Then

Fix-points of Meromorphic Functions

where Aj

=1=

101

0 (j = 1,2) are constants. It follows that

P2(Z) = Al {P(z) - zdd + P(z) = Al {AI(Z - ZI)d + z - zdd + Adz - zdd + z , P2(Z) - Z = Adz - zdd{[AI(Z - zdd-I + 1jd + 1} . (2.88) On the other hand, we have

(2.89) The right members of (2.88) and (2.89) are polynomials of Z-ZI. Since these two polynomials do not have the same coefficients, we get a contradiction. Since the polynomial P(z) -z has zeros, the polynomial P(z) always has fix-points. As in the case of transcendental entire functions, a fix-point Zo of P(z) is said to be attractive, repulsive or neutral, according to IP'(zo) I is less than 1, greater than 1 or equal to 1. In the third case, we have P'(zo) = e2'1rOi, (0 :5 9 < 1); we distinguish again two cases according to the number 9 is rational or irrational. Moreover for these three kinds of fix-points of P(z), we have correspondingly four theorems which are obtained from Theorems 2.20, 2.21, 2.22 and 2.23, in replacing J(z) by P(z). Now we are going to prove the following theorem:

Theorem 2.29. There exists a number R > 1 such that in the domain Izl > R, the sequence IPn(z)l(n = 1,2, ... ) converges uniformly to 00. Proof. We have Since d ~ 2, we can find a number R > 1 such that for Izl > R, we have

Then for Izl > R, we have

IP(z)1 > 21z1 > 2R > R , IP2(z)1 > 2IP(z)1 > 221z1 > 22R > R, IP3(Z)1 > 2IP2(z)1 > 231z1 > 23R > R , in general

Fix-points and Factorization of Meromorphic Functions

102

Hence in the domain Izl > R, the sequence IPn(z) I (n = 1,2, ... ) converges uniformly to 00. Theorem 2.29 shows that the point z = 00 is one at which the family {Pn(z)(n = 1,2, ... )} is normal. On the other hand, the point z = 00 is a fix-point of P(z) and it is attractive in the sense that the point w = 0 is an attractive fix-point of the function

wd

1

p(t) =

aowd+alwd-l+ ... +ad_lw+ad

Finally we are going to prove the following theorem:

Theorem 2.30. The set J(P) has the following properties: 10 J(P) is a non-empty bounded perfect set_ 2 0 Let En be the set of the fix-points of Pn(z) and define 00

E(P) =

U En . n=l

Then each point of J(P) is a point of accumulation of E(P).

Proof. First we prove that J(P) is non-empty. In fact, if J(P) is empty, the family {Pn(z)(n = 1,2, ... )} is normal in the complex plane. Let Zo be a fix-point of P(z). We have

Pn (zo) = Zo

(n = 1, 2, ... ) .

So the sequence Pn(zo)(n = 1,2, ... ) is bounded. Then by Lemma 2.9, the sequence Pn(z)(n = 1,2, ... ) is locally uniformly bounded in the complex plane. This contradicts Theorem 2_29_ By Theorem 2.29, the set J(P) belongs to the circle Izl ~ R and therefore is bounded. To prove that J(P) is perfect, it is sufficient to show that J(P) c {J(P)}'. First of all, by Theorem 2.13, there exist three positive integers n. (i = 1,2,3) such that nl < n2 < n3 and that there are three points ~.(i = 1,2,3) which are fix points of P(z) of exact order n.(i = 1,2,3) respectively. Obviously ~.(i = 1,2,3) are distinct. Since Pnj(~.) = ~.(i = 1,2,3), we see that the sequences Pn(~.)(n=I,2,

are bounded.

... ) (i=I,2,3)

(2.90)

103

Fix-points of Merom orph ic Functions

Now consider a point Zo E J(P) and a circle c : going to show that in c there is a point z' such that

z' i= Zo, z' E

Iz - zol <

J(P) .

r. We are

(2.91)

This is proved by the same method used in the proof of the part 2° in Theorem 2.24, with suitable modifications. Since the family {Pn(Z)( n = 1,2, ... )} is not normal in c, there is a subsequence Pnk (z)(k = 1,2, ... ) such that from the sequence Pnk (z)(k = 1,2, ... ) we cannot extract a subsequence which converges locally uniformly in c to a holomorphic function or to the constant 00. Consider the sequence Pnk (zo)(k = 1,2, ... ). From this sequence we can extract a subsequence Pm" (zo)(h = 1,2, ... ) converging to a limit >., finite or infinite. Distinguish two cases: 1) >. is finite. Among the three points ~i (i = 1,2,3) there are at least two points, say ~1 and ~2' not equal to >.. Choose two bounded domains OJ (j = 1,2) satisfying the following conditions: a) ~i E 0i (i = 1, 2), 0i n D i= tP (i = 1, 2) where D is the domain Izl > R in Theorem 2.29. b) 0 1 n O 2 = tP, >. f/: Oi (i = 1,2) where Oi denotes the closure of Oi

y

x

Then evidently we can get two positive numbers M and 5 such that for any two points Zi E Odi = 1,2) we have (2.92)

104

Fix·points and Factorization of Merom orphic Functions

and that

(2.93) where "f denotes the circle Iz - >'1 < o. In view of the condition a) and the boundedness of the sequences (2.90), we see that the family {Pn(z)(n = 1,2, ... )} is not normal in 0 •. Hence there are two points (\(.(i = 1,2) such that (\(. E 0. (i = 1,2), (\(. E J(P) (i = 1,2) . (2.94) Let H be a positive integer such that

h~ H .

Pm" (zo) E"f for

(2.95)

As in the proof of the part 2° in Theorem 2.24, we see that we can get an integer h ~ H such that the values taken by the function Pm" (z) in the circle c cover one of the domains 0. (i = 1, 2), say 0 1 . So in c there is a point z' such that (2.96) Then by (2.94) (\(1

E

J(Pm ,,),

z'

E

J(Pm ,,) = J(P)

and by (2.93)' (2.95)' (2.96)' we have z'

I- Zo

.

Hence the point z' satisfies the condition (2.91). 2) >. is infinite. Let H be a positive integer such that

IPm,,(zo)1

> M for h

~ H

(2.97)

where M is the number in (2.92). As above, we can get an integer h ~ H such that the values taken by the function Pm" (z) in the circle c cover one of the domains O.(i = 1,2). By (2.97), again we see that in c there is a point z' satisfying the condition (2.91). It remains to prove the part 2° of Theorem 2.30. Let ai(j = 1,2, ... , q) be all the distinct roots of the equation P'(z) = D. Consider a point Zo E J(P). Since Zo is a point of accumulation of J(P), for any circle c : Iz-zo I < r, there is a point z' E c such that

z'

E

J(P), z' I-

Zo,

z' I- P(ai) (j = 1,2, ... ,q) .

Fix-points of Meromorphic Functions

Let a be a root of the equation P(z)

P(a) = z',

= z'.

105

Then

P'(a) =I 0 .

Next as in the proof of the part 3° of Theorem 2.24, we see that there is a point Zl E c such that Zl

=I Zo,

Zl

E

E(P) .

Hence Zo is a point of accumulation of E(P). For further study of the Julia sets of polynomials and rational functions, the reader is referred to the following works: 1. H. Brolin, "Invariant sets under iteration of rational functions", Ark. Mat. 6 (1965) 103-144. 2. P. Blanchard, "Complex analytic dynamics on the Riemann sphere" , Bull. American Math. Soc. 11 (1984) 85-141.

3 FACTORIZATION OF MEROMORPHIC FUNCTIONS

3.1. INTRODUCTION The factorization theory of meromorphic functions concerns whether a given function can be expressed as a composition of two or more nonlinear meromorphic functions. This theory was developed about two decades ago. The investigation is closely related to the study of the fix-points of a function. A complex number Zo is said to be a fix-point of J{z) iff J{zo) = zoo As far back as 1926, Fatou claimed that for any nonlinear entire function J, the iteration J(f)( = h) has at least one fix-point. This fact was formally proved in 1952 by P.C. Rosenbloom utilizing Picard's theorem. The proof assumes if J(f) has no fix-point at all, then clearly, it is not possible for J(z) to have any fix-point either. Therefore, the function

J{z) - z F{z) = J(f(z)) - J{z) is entire and assumes neither 0 nor 1. According to Picard's theorem F must be a constant, say c. Clearly c =1= 1. From the above equation, we get

c[J(f(z)) - z] = J{z) - z . By differentiating both sides, we have

c[J'(f{z))j'(z) - 1] = j'(z) - 1 107

108

Fix-points and Factorization of Merom orphic Functions

or

!,(z)[c!,(f(z)) - 1] = c - 1 # 0 . It follows that J'(z) never vanishes, and moreover, J'(z) never takes the value ~. Thus J' has to be a constant, hence J is linear. This contradicts the assumption and therefore proves the assertion that J(f) must have infinitely many fix-points for any nonlinear entire function f. In the same paper, Rosenbloom extended this result and obtained two theorems (given below) by applying the newly developed Nevanlinna's valuedistribution theory. Since then, the value-distribution theory has greatly affected the research in the factorization theory.

Theorem 3.1. Let J and g be two transcendental entire functions. Then either J or J(g) must have infinitely many fix-points. Theorem 3.2. Let P(z) be a nonlinear polynomial and J(z) be a transcendental entire function. Then P(f(z)) must have infinitely many fix-points. It was in the same paper that Rosenbloom first introduced the concept of "prime function". He defined an entire function F(z) to be prime if every factorization of the form F(z) = J(g(z))(= J 0 g(z)). J, g being entire implies that one of the functions, J or g, must be linear. Rosenbloom asserted, without giving a proof, that eZ + z is a prime function and remarked that its proof was quite complicated. Given the present techniques in the study of factorization, the proof that eZ + z is prime is a relatively simple matter. It was not until 1968 that F. Gross gave a broader definition of the factorization for meromorphic functions. He not only provided a proof of the primeness of eZ + z but also started a series studies on factorization theory. In this book, the emphasis will be on the development of the methods for testing whether a given meromorphic function can be factorized as two or more nonbilinear meromorphic functions. More specifically, we shall discuss (i) the forms of the factors in a factorization; (ii) the existence of fix-points and the factorization; (iii) the criteria for pseudo-prime functions; (iv) the growth rates of meromorphic solutions of certain functional equations and (v) the factorizations of meromorphic solutions of linear differential equations and the uniquely factorizability of certain functions.

Factorization of Meromorphic Functions

109

Through the investigations of Gross and Yang in the U.S.A., Goldberg and Prokopovich in the U.S.S.R., Baker and Goldstein in England, Steinmetz in Germany, and Ozawa, Urabe and Noda in Japanj the theory of factorization has become a new branch in the value-distribution theory of meromorphic function. For the interest of the reader, we have included many open questions and research problems for further studies in this book. 3.2. BASIC CONCEPTS AND DEFINITIONS

Definition 3.1. Let F(z) be a meromorphic function. If F(z) can be expressed as (3.1) F(z) = f(g(z))(= f 0 g(z)) , where f is meromorphic and 9 is entire (g may be meromorphic when f is rational)' then we call expression (3.1) a factorization of F (or simply a factorization), and f and 9 are called the left and right factors of F, respectively.

Definition 3.2. If every factorization of F of the above form, implies that either f or 9 is bilinear (J is rational or 9 is a polynomial), then F is called prime (pseudo-prime). Definition 3.3. If every factorization of the form (3.1) leads to the conclusion that f must be a bilinear form when 9 is transcendental (or f is transcendental and 9 must be linear) then F is called left-prime (rightprime). When factors are restricted to entire functions, it is called a factorization in entire sense. Under such a provision a prime (pseudo-) function will be denoted as E-prime (E-pseudo-prime). Nevertheless, we shall prove in the sequel that if a nonperiodic entire function F is E-prime (E-pseudoprime) then it also must be prime (pseudo-prime). In other words, we only need to consider entire factors for the primeness (or pseudo-primeness) of a non periodic entire function. Until recently the majority of the research accomplishments in the factorization theory have been based on the studies of the prototype ele + Zj the construction of certain families of prime or pseudo-prime functionsj the finding of sufficient conditions for a certain class of functions being prime or pseudo-prime, and the discussions of problems of the uniquely factorizability of certain functions as well as the commutativity of factors. In all these investigations, the Nevanlinna value-distribution theory has been

110

Fix-points and Factonzation of Merom orphic Functions

used as the primary tool. Also, in the development of the proofs, the following properties of meromorphic functions have been used: (i) the growth property; (ii) the distribution of the zeros or the existence of defect values; (iii) the periodicity; (iv) the fix-points and (v) being a solution of a linear differential equation. Generally speaking, the research in the factorization theory is still in its infancy stage. There are many interesting questions to be studied and resolved. We strongly believe that value-distribution theory can be further perfected by studying the factorization theory. Here we shall only deal with factorization of transcendental meromorphic functions, since Ritt has obtained a complete theory on the factorization of polynomials (but not rational functions!). 3.3. FACTORIZATION OF CERTAIN FUNCTIONS In this section we shall prove that for any non-constant polynomial P(z)e Z + P(z) is prime; a generalized form of eZ + z. Prior to this proof it is natural for us to ask: For a given function, is there any link between the forms of the factors and that of the given function? More precisely, we ask whether certain classes of entire functions and their factors possess more or less similar properties? The answer is "yes". Before proceeding further, we introduce some definitions and lemmas below:

Definition 3.4. We shall call an entire function F(z) periodic mod 9 with period T, if and only if the following identity holds:

F(z + T) - F(z) = g(z) . Sometimes, we simply call such a function a pseudo-periodic function mod g. For instance given the function F(z) == eZ + P(z), F is periodic mod a polynomial with period 21ri.

Theorem 3.3. Let F(z) be an entire function and periodic mod a polynomial P(z) with period T. If F = fog, then 9 must assume the following form: (3.2) where Hi(Z),i = 1,2, are periodic functions with the same period constant, and q(z) is a polynomial.

T,

c is a

111

Factorizatzon of Meromorphic Functions

Proof. We may assume, without loss of generality (w.l.o.g) that According to the above hypothesis we have

T

=

1.

J(g(z + 1)) - J(g(z)) = P(z) . We note that whenever g(zo + 1) = g(zo) for some Zo, the function on the left side of the above equation will assume a value zero. Therefore, we must have g(z + 1) - g(z) = PI(z)e adz ) , (3.3) where Pdz) is a polynomial and adz) is an entire function. Similarly, we deduce (3.4) where P2(Z) is a polynomial and a2(z) is an entire function. Substituting z with z + 1 in equation (3.3) and then adding to equation (3.4) we get

g(z + 2) - g(z) = Pdz + l)e adz +l) + Pdz)ea.(z) .

(3.5)

Equating (3.4) and (3.5) we get

Applying Borel's lemma, we conclude that

adz

+ 1) = adz) + c ,

c a constant. We write

Then

Thus H2(Z) is a periodic function with period 1. We easily verify (for instance, by equating the coefficients) that for any given polynomial Pdz) there always exists a polynomial q(z) such that

eCq(z + 1) - q(z) = Pdz) .

112

Fix-points and Factorization of Meromorphic Functions

Setting

Hdz) = g(z) - q(z)e adz ) = g(z) - q(z)e H2 (z)+cz ,

(3.6)

we obtain

Hdz + 1)

g(z + 1) - Pdz + 1)e H2 (Z+I)+CZ+C = g(z) + q(z)e H2 (z)+cz - eCq(z + 1)e H2 (z)+cz = g(z) - (eCq(z + 1) - q(z))e H2 (z)+cz = g(z) - PI (z)e H2 (z)+cz = HI(z) .

=

Hence HI (z) is also a periodic function with period 1. It follows from (3.6) that and the theorem is thus proved.

Remark. (i) H F is periodic mod a nonconstant entire function h(z) with p(h) ~ 1, then the theorem remains valid, where q(z) need not be a polynomial, but p(q) ~ 1. (ii) In general, it seems there is not much that we can really say about the factors I and g in a factorizatio~ F = I(g). However, for functions F of certain forms, we can determine the possible forms of I or g in the factorization F = I(g). Gross, Koont and Yang proved: Given F(z) == HI(Z) + ze H2 (z), where HI and H2 are periodic entire functions with the same period T. For any factorization F(z) = I(g(z)), where I, g are entire functions with I being nonlinear, then g(z) must be of the form g(z) = T(z) + az, where T(z) is a periodic entire function with period T and a is a non-zero constant. Furthermore they also proved if H2 is prime, so is F. Theorem 3.4. H P(z) is any nonlinear polynomial and g(z) is an arbitrary transcendental entire function, then P(g) is not periodic mod a non-constant polynomial. Proof. Assume the theorem to be false; then there exists a non-constant polynomiall(z) such that

P(g(z + 1)) - P(g(z)) By the above theorem, g is of the form

=

l(z) .

(3.7)

Factorization of Meromorphic Functions

113

where Hi(Z),i = 1,2 are periodic entire functions with period 1, q(z) is a polynomial, and c is a constant. Substituting z by z + 1, z + 2, ... ,z + n - 1 successively in (3.7) and adding them up, we obtain

P(g(z + n)) - P(g(z)) = l(z)

+ l(z + 1) + ... + l(z + n -

1) .

(3.8)

Alternatively, from (3.7) we can derive

g(z + n) = g(z)

+ [q(z + n)e cn - q(z)Je H2 {z)+cz

•

Substituting this into (3.8), we get

P(g(z) + [q(z + n)e cn - q(z)]e H2 {z}+cz) = P(g(z)) + l(z) + ... + l(z + n -1). (3.9) IT leci < 1, then at z = 0, the right side of (3.9) tends to infinity with n, whereas the left side is bounded. This is a contradiction. When leci > 1, replacing z by -z in (3.9), we will arrive at a similar contradiction. Hence we may assume that leci = 1. Let t be the degree of l(z). When z = 0 and n is sufficiently large, then the absolute value of the right side of (3.9) is greater than

but less than where .Al,.A2 are suitable constants. Assume the degree of P = u and the degree of q = v then from (3.9) we conclude that uv = t + 1. Now we shall treat the two cases separately: (a) a(z) = eH2 (z)+cz = a, a constant and (b) a(z) is not a constant. IT case (a) holds, then

P(H1 (z)

+ aq(z + 1)) - P(Hl(Z) + aq(z)) = l(z) .

For large r(= Izl) the maximum modulus of the left side of the equation is greater than .A(M(r, Ht})u-l for some positive constant .A. This is impossible, since u 2:: 2 and HI (z) is transcendental. Now assume that case (b) holds. Then it can be easily verified that for every z the right side of (3.9) is less in absolute value than .Ant+l for some positive constant .A (independent of z) and sufficiently large n > N(z)j N(z) is a quantity that depends on z. In order to estimate the left side of

114

Fix-points and Factorization of Meromorphic Functions

(3.9), we may assume without loss of generality that P(z) and q(z) assume respectively the following forms:

P(z) = >'uzu + ... q(z)=zt/+ ... We can choose Zo so that

Then for any 0 < e < 1 and sufficiently large n (depending on N(zo) and e), the left side of (3.9) is greater in absolute value than

(3.11) It follows from this and (3.10) that

>'nt+1 > (1 - e)2>.nt+ 1

,

for sufficiently large n and any small e. This will lead to a contradiction and the theorem is thus proved.

Remark. The theorem remains valid if the order of 9 is assumed to be less than 1, then the function q( z) in the expression of g( z) will satisfy

p(q) ::; 1. Theorem 3.5. If P(z) is a polynomial of degree> 2 and if f is a transcendental entire function, then f(P) is not periodic mod q (q a polynomial). Proof. Suppose that the theorem is not true. We may then assume that the function F(z) == f(P(z)) has a pseudo-period i such that

F(z + i) - F(z) =
q(z + 1) - q(z) =
M(r, F) >

r2 M(2r,


(3.12)

115

Factorization of Meromorphic Functions

Also for any positive integer n,

F(z + in) - F(z)

=


(3.13)

From the above two equations we can derive for every e > 0, given sufficiently large r > R(e) and Inl < r:

(1 - e)M(r, F) < M(r, F(z

+ ni)) < (1 + e)M(r, F) .

Let zo(lzol = r) be a point such that IF(zo) I = M(r,F). Recalling the degree of P(z) 2: 3, we can easily see that, for sufficiently large rand sufficiently small 6, there exists a point z' (i- zo) in the sector 6 < arg z < 'II" - 6 such that

Iz'l Izol

-

"" 1

and

P(zo) = P(z') .

Thus for n > 0,

M(r, F)

= If(P(z'))1 = IF(z')1 = IF(z' + nil -
Using simple geometric illustration we can choose n so that z' + ni = x + iy with Iyl < (when n < 0, the form (3.13) will be changed slightly, but the proof is not affected). Using inequality (3.12), we obtain from the above equation that

!

2

M(r, F) < M(x + 1, F) + 2rM(2r,
or

(1- ~) M(r, F) < M(x + 1, F) = M(1 + Iz'l

cos 6, F)

< M(>..r, F)

(3.14)

for some >.. satisfying cos 6 < >.. < 1. On the other hand by the Hadamard Three-circles theorem (choosing rl = 1, r2 = >"r, and r3 = r) we have

M(>..r,F) ::; kM(r,F)l-logt!IOgr,

k a positive constant.

If follows from the above equation that for any given positive e < 1 we have

M(>"r, F) < eM(r, F) for sufficiently large r.

(3.15)

116

Fix-points and Factorization of Meromorphic Functions

Therefore Eqs. (3.14) and (3.15) lead to

Since this is impossible, the theorem is proven.

Remark. Whittaker proved that for any given entire function 9 and constant c, the functional equation

cf(z + 1) - f(z) = g(z) has an entire solution f(z) satisfying p(f) = p(g) if c = 1. When c i- 1 and p(g) < 1, Gross observed that the above functional equation has an entire solution f with p(f) :::; 1. In addition he asserted that if p(f(P)) ~ 1, then f(P) is not periodic mod any function of order less than one. Using the above theorems we can derive the following theorem.

Theorem 3.6. Let F be an entire function of exponential type and periodic mod some non-constant polynomial. Then F is either prime or of the form

F(z) = f((z + c)2) , where f is an entire function and c is a constant. Before proceeding with the proof, we need two additional results which are interesting in their own right.

Lemma 3.1. Let F be a non-periodic transcendental entire function. Then F is prime if and only if it is E-prime. Proof. Assume that F is E-prime but not prime. Then there must be a factorization of the form

F=fog, where f is a meromorphic (not entire) function. If f is transcendental, then f must be of the form

where n is a positive integer and the form,

g(z)

= S-o

+ s-a(z) ,

It is an entire function. Then 9 must have where a an entire function.

Factorization of Meromorphic Functions

117

Thus

F(z)

= 10 g(z) = e-na(z) h(~o + ea(z)) = [e- nr h(~o + ~)l a(z) 0

.

Since the left factor, e- nr I(~o+d is a nonlinear entire function, and assuming that F is E-prime, we conclude that a(z) must be linear. This implies that 9 is periodic and so is F. This contradicts the hypothesis that F is not a periodic function. When I is rational (but not a polynomial), and using a similar argument to the one used in the preceding case, the same contradiction will be reached. This completes the proof of the lemma. Equipped with this lemma, for the problem of factorizing non-periodic functions such as eZ + P(z) (where P(z) is a non-constant polynomial) we shall only need consider the entire factors. To prove that a transcendental entire function F is E-prime a general approach will be (i) to prove that F is pseudo-prime, and (ii) to show that when F = loP or F = Pol with P being a polynomial, P has to be linear.

Lemma 3.2. Suppose that F is an exponential-type entire function

F(z + r) - F(z) = P(z)e az , where P(z) is a polynomial and a is a constant. Then F must be pseudoprime.

Proof. Assume that F entire. Then

= log

and that

I, 9 both are transcendental

I(z(z + r)) - I(g(z)) = P(z)e az . Thus all the zeros of g(z

(3.16)

+ r) - g(z) are among the zeros of P(z). Hence,

g(z + r) - g(z) = q(z)e CZ

,

(3.17)

where q is a polynomial and c is a constant. It is easy to conclude from the growth rate of a composite function that 9 is at most of order 1 and of minimum type. Consequently if q(z) is not identically zero, then the constant c in (3.17) has to be zero. In any case, there exists a positive integer n that is sufficiently large so that g(n) (z + r) - g(n) (z) = O. If g(n) fails to be a constant, then it is a periodic function and has at least a growth

118

Fix-points and Factorization of Meromorphic FUllctions

of order 1 and of finite type. This is a contradiction. Since g(n) (z) has to be a constant, g(z) is a polynomial. This also proves that F is pseudo-prime. We can now go back to the proof of Theorem 3.6.

Proof of Theorem 3.6. By Lemma 3.2, if F is not prime then F has only two possible factorizations: (a) F = loP or (b) F = Po Ij I is transcendental and p is a nonlinear entire polynomial. However, according to Theorem 3.4, the factorization of form (b) can be ruled out. Finally, according to Theorem 3.5, F can only be factorized as F = loP when P is a polynomial of degree two. The theorem is thus proved. Using the preceding theorem, we can immediately obtain the next one that was mentioned in the beginning of this chapter.

Theorem 3.1. Let P(z) be any non-constant polynomial, and let 0), b be any two constants. Then F(z) = eaz+b + P(z) is prime.

a(~

Proof. Using the last theorem, we only need to show that it is impossible to express eaz+b + P(z) as I((z + c)2)j I entire and c a constant. The proof of this simple exercise is left to the reader.

Remark. If in the theorem it is assumed that P is an entire function of order less than 1, then the conclusion remains valid. Bascially we have proved that e Z + z is prime. Ozawa proved that ee" + z is prime. Ozawa, Yang, Gross, and Urabe have independently proven that for any positive integer, en(z) +z is primej en(z) is the n-th iteration of eZ , i.e., en(z) = en_de Z ). These naturally lead to the question: For any given non-constant periodic entire function H, must eH + z be prime? Or more generally, must H(z) + z be prime? No definite answer has been found for the former question (as far as we know of). For the latter one (in general it is not true!) we have the following result.

Theorem 3.8. Let H be a periodic entire function of finite lower order, and a be a nonzero constant. Then H(z)

+ az

is prime.

Discussion. If H is of infinite lower order, will the above theorem remain true? The answer is no. However, if in theorem 3.7, the polynomial P is a constant, then clearly F becomes a periodic entire function and is not prime. Gross raised this question: Does there exist a periodic entire function that is also a prime function? With regards to this question, Ozawa first provided some solutions that are E-prime. Later on, Yang and Gross

Factorization of Meromorphic Functions

119

also obtained some general results that are related to the above question. Ozawa exhibited some E-prime periodic entire functions of order 1 and infinite. We state the following result that seems to be a most general one in resolving Gross' question.

Theorem 3.9. (Yang and Gross) Let (3 be an arbitrary non-constant entire function, and P be a non-constant polynomial. Then

F(z) = (e Z

-

1) exp(P(e- Z )

+ (3(e

Z ))

is E-prime. The reader is referred to Gross and Yang's paper "On prime periodic entire functions", Math. Zeit 1'14 (1980) 43-48, for the proof. Note that in the text of the above paper, only the E-primeness of F has been established. It remains an open question whether there exists a periodic entire function which is also prime. 3.4. FACTORIZATION OF FUNCTIONS IN COSINE OR EXPONENTIAL FORMS In this section we explore the possible forms of the factors in the factorization of functions in cosine or exponential forms. We first state (without giving a proof) the following lemma:

Lemma 3.3. (Edrei). Let g(z) be an entire function. H there exists an unbounded sequence of numbers {an} such that all, except finitely many of the values n, are the roots of the equations:

g(z) = an, n = 1,2, ... , and lie on the same straight line, then g(z) must be a polynomial with its degree not greater than 2. It was then conjectured by Ozawa that if there is a sequence {an} such that lanl -+ 00 and that all but except a finite number of roots of J(z) = an lie on p straight lines: h, l2,'" ,lp, any two of them are not parallel to each other, then J(z) reduces to a polynomial of degree at most 2p. Consequently, J. Qiao proved this conjectured for p = 2,3 and recently, F. Ren presented a proof for any positive integer p ~ 4. As a consequence of these, one can show immediately that if J and 9 are two finite order entire functions with all but a finite number of their roots lie on finite

Fix-points and Factorization of Meromorphic Functions

120

straight lines, then the product I(z)g(z) is pseudo-prime. For instance, sin z sin ".fiz sin v'3z are pseudo-prime.

Lemma 3.4. Let I be an entire function of order less than ~, and g be an arbitrary entire function. A necessary and sufficient condition for F = log being periodic is that g is a periodic function. Proof. Assume that F has a period T. Choose a point Zo and avoid a (= F(zo)) a branch point of the inverse function 1-1. Consider the straight line L : s" = Zo + tT, -00 < t < 00. Then clearly F is bounded on L. We are going to show that g is also bounded on L. IT g is unbounded on L, then g(L), the image of Lunder g, is a path tending to infinity on which I is bounded. Since the order of I is < ~, by Wiman's theorem this is impossible. Therefore, {g(zo + nT)}n=1 is bounded. Hence, there exists a constant k > 0 such that

Ig(zo

+ nT)1

~ k

'
We also have for n = 1,2, ...

I(g(zo + nT)) = I(g(zo)) = a . Thus all the points in "In = g(zo + nT) are among the finite roots of the equation 1('1) = a that lie in I'll ~ k. Hence, there must exist some m =t n such that g(zo + mT) = g(zo + nT). Moreover, for sufficiently small €,

I(g(zo + € + mT))

= F(zo + € + mT) = F(zo + € + nT)

= I(g(zo

Now since

I

+ € + nT)) .

is one-one in a sufficiently small region around the point

g(zo + € + nT) (by choosing a = I(g(zo))), we must have g(zo + € + mr) = g(zo + € + nT) for all sufficiently small €. Hence g(zo + mT) == g(zo + nr), and g has period (m - n)T. This completes the proof of the lemma. Theorem 3.10. The function F(z) = cosz is pseudo-prime. Furtherthe possible forms of the factorization of F = log are as follows: I(s-) = cos~, g(z) = z2 j

more, (i) (ii) where

I(s-) = Tn(s-), g(z) = cos~, Tn(s-) denotes the nth Chebyshev polynomial (n 2: 2),

Factorization of Meromorphic Functions

(iii) f(d

= ~(~-n + ~n), g(z) = eiz/nj n denotes

121

a non-negative integer.

Proof. The E-pseudo-primeness of cos z follows immediately from Lemma 3.2. We now show that cos z is also pseudo-prime. Assume that cosz = f 0 (g(z)), where J(~) is a transcendental (non-entire) function, and 9 is transcendental entire. Then by P6lya's theorem p(!) = 0, and, moreover, f has at most one pole. Hence, J(~) = (~- ~o)-n fdd for n ~ 1j It is an entire function with fd~o) t= 0, and g(z) = ~o + eM(z)j M is a non-constant entire function. Thus cos z

= It (g(z))e-nM(z)

Let {'7i} be the zero set of fd~) that is an unbounded set. To each '7i, all the roots of the equation g(z) = '7i are real since cos z has only real zeros. By Lemma 3.2, we conclude that g(z) is a polynomial of degree no greater than 2. This is a contradiction. Hence, cos z is pseudo-prime. Now we investigate the cases where the factors are restricted to entire functions only. (a) cos z = f 0 g(z)j f entire, 9 a polynomial. By the above argument, it follows that 9 is a polynomial of degree no greater than 2. Then 9 can be expressed as

g(z) = k(z - a)2 + bj

k

t= 0,

a, b are all constants.

This is the form (i) mentioned in Theorem 3.lD. (b) cos z = Pn 0 g(z), where Pn is a polynomial of degree ~ 2, and 9 is an entire function. By Lemma 3.4, we conclude that 9 is a periodic function with period, say T. We have cos(z + T)

= Pn(g(z + T)) = Pn(g(z)) = cos z ,

(3.18)

then, T = 2l7r for some integer l( t= 0) and g(z) is a function of an exponential type and is periodic with period 2l7r. It can be expressed as m

g(z) =

L

ak

eikz / l

•

k=-m

Substituting this into (3.18) and applying Borel's lemma, we can conclude l = n, m = 1. Then g(z) can be written as

122

Fix-points and Factonzation of Meromorphic Functions

We may assume without loss of generality that ao

= 0, a1 =

~, and can set

Using the above two expressions and (3.18), we have 1 0 _elZ 2

1 IZ 0 = Po + _e2

n

(1_elZ 0 /n 2

+ a_

0/ n ) e- IZ

1

(3.19)

Equating the coefficients of eiz and e- iz respectively on both sides of (3.19)' we get An = 2n - 1 and a~1 = 2- n . Hence, a -1

=

e201ri/n ,

8

= 0, 1,2 ...

,n - 1 .

And consequently,

= e'''''0/ n

z-

81r

cos - - - , n

(8

= 0, 1,2, ... ,n - 1) .

Let Tn(w) denote the n-th Chebyshev polynomial, namely

Tn (cos z) = cos nz .

(3.20)

Equations (3.19) and (3.20) give us,

In fact,

Setting 8 = 0, we get g(z) = cosz/n and Pn (77) = Tn(77). Thus, to a fixed n, all the factorizations of Pn,. 0 (gn,.)(8 = 1,2, ... n - 1) are equivalent to Pn,o 0 gn,o, that is the desired form (ii) of the theorem. (c) Finally, let us assume cosz = f 0 g(z), where f is rational (but not a polynomial) and 9 is transcendental entire. Then as before we have,

123

Factorization of Meromorphic Functions

where n ~ 1, P is a polynomial of degree t, and a(z) is a non-constant entire function. Obviously the function a has to be linear, namely a(z) = az + b; a i- 0 and b is a constant. Thus cos z

=

P(~

+ eaz+b) exp{ -n( az + b)} = o.

We may, without loss of generality, assume that above two expressions that

~o

where at, at-l, ... ,ao are suitable constants; at Borel's lemma, we immediately obtain

i- o.

. It follows from the

By an application of

t = 2n, at-l = ... = al = 0 , and

or a

i l l at = -2"' ao = -2"

=~,

Form (iii) is now established, and the theorem is proven.

Remark. Another justification that we shall learn about in the sequel (Chap. 4) for cos z being pseudo-prime is that it satisfies an ordinary linear differential equation: y"(Z) - y(z) = o. Theorem 3.10 shows that for any given positive integer k, cos z can be factorized as (3.21) where Pk is a suitable polynomial of degree k, and gk is a corresponding entire function. Ozawa established a converse to Theorem 3.10 as follows.

Theorem 3.11. Let F be a non-constant entire function. Suppose that for each integer k = 2i(J' = 1,2, ... ) and k = 3 or 5, (3.22) holds for some polynomials Pk and entire function

F{z) =

aeH(z)

+b

gk.

Then either

(3.23)

124

Fix-points and Factorization of Meromorphic Functions

or

+b,

F{z) = acos y'H{z)

(3.24)

where a{,e 0) and b are constants and H{z) is an entire function.

Remark. Ozawa also proved that the conclusions made in Theorem 3.11 remain valid under weaker hypotheses; namely (3.22) holds for k = 3j (j = 1,2, ... ) and k = 2,4. Ozawa's proof is rather complicated. Recently, J. Huang and G.D. Song improved Ozawa's method and obtained the following stronger result with a simplified argument.

Theorem 3.12. Let F{z) be a non-constant entire function. Suppose that (3.22) holds for each integer k = q (a given positive integer) k = nj with 2 ~ nl < q, (nj, q) = 1, njH ~ njq, j = 1,2, .... Then F{z) has either form (3.23) or form (3.24). Lemma 3.5. Let Q(d be a given polynomial of degree p. Then there exist only finitely many pairs of linear functions U and V such that

VoQ=QoU, unless Q(d = A(S- - alP

+ B, A, B

are constants.

Proof. Let V{t) = at + b, U(t) = et + d. If V

0

Q = aQ(s-)

+b= Q0

U

= Q(es- + d) ,

then, by differentiating, we have

aQ'(d

=

eQ'(es- + d) .

This shows that the zero set Z of Q' is invariant under mapping: S- -+ es- + d. Moreover e,e 1 (since Z is a finite set, unless V(S-) = U(d = S-). Then by setting a = d/{1 - e), we have

U(t) = e{t - a)

+a

.

Therefore, unless Z = {a}, the invariant property of Z leads to the conclusion that lei = 1. If Z ,e {a}, then c is a root of unity, and so we may assume eN = 1, but ek ,e 1(0 < k < N). Thus Z consists of the vertices

125

Factorization of Meromorphic Functions

of a regular N -polygon centered at the point a and possibly the point a itself. It follows that M

Q'(d = c(s- - a)"

II {(s- -

a)N - b;} .

;=1 By integrating, we get

where h is a polynomial and B is a constant. In the meantime, we also have and thus

It is easy to see from the above two expressions that there can be only finitely many such V(t), so the number of the function U(t) is also finite. Now, suppose that Z = {a}, then

Q'(s"} = c(s- - a)p-l,

Q(d = A(S- - alP

+B

.

Under this circumstance, it is obvious that there are infinitely many pairs of U, V with U(t) = c(t - a) + a, V(t) = cPt + B(l - ct ) and satisfying V 0 Q = Q 0 U. Therefore Lemma 3.5 is proven. Lemma 3.6. Let J(s) and g(z) be two non-constant entire functions. Let Pm and P n be two polynomials of relatively prime degrees m and n, respectively. If for all z which belong to the complex plane the following identity holds: Pm 0 J(z) = Pn 0 g(z) , then there exist entire function s(z), and polynomials U(d and V(d of degree nand m, respectively such that

J(z) = U(s(z)), Proof.

g(z) = V(s(z)) .

We first factorize Pm(u) - Pn(v) into irreducible factors in

C[u, v]. Let R(u, v) be such a factor, then we have

R(f(z), g(z)) = 0 .

126

Fix-points and Factorization of Meromorphic Functions

Using a theorem of Picard's that can be phrased as: if R(u, tI) is an irreducible polynomial in C[u, til and there exist some non-constant entire functions J(z) and g(z) satisfying R(f(z), g(z)) = 0, for all z, then the Riemann surface of R(u, tI) = 0 has genus 0 and will be denoted by S (conformationally equivalent to the Riemann sphere). If J and gin R(f(z), g(z)) = 0 can be meromorphic then the genus of such a surface X is less than or equal to 1. Moreover, any Riemann surface of genus 1 can only be uniformized by elliptic functions not by entire functions, and there exists a conformal map t/J that maps the points p on the Riemann surface R (defined by R(u, tI) = 0) onto points ~ of the Riemann sphere; i.e., ~ = t/J(p). We may assume without loss of generality that ~ = 00 corresponds to a point with u = 00. Except at a finite numbering branch points of R we may use u as a local uniform parameter. Thus, except at those branch points, ~ is a holomorphic function O"(u) of u E R. Therefore, the map

z -+ (f(z), g(z)) = p -+

S

= t/J(p) = 0" 0 J(z) = s(z)

is holomorphic at all z E R, except perhaps those for which (f(z*), g(z*)) = (u* , tI*) = p* is a branch point of R. Clearly these exceptional points z* form a discrete set E. Also if z tends to z* E E this implies s(z) tends to s(z*). Therefore, z* is a removable singularity of s(z), and hence s(z) is entire. Now the Riemann surface S, conform ally equivalent to the Riemann sphere, is defined by R( u, tI) = 0 and has genus zero. Then there exist uniformizable functions

u=U(d,

tI=V(d,

where U, V are rational functions such that a 1- 1 correspondence between points of S and the Riemann sphere is established. It follows that there exists u = U(d and tI = V(~) where U, V are rational functions on the surface R mentioned before, such that

J(z) = U(s(z)) and g(z) = V(s(z)) .

(3.25)

Suppose R(u, tI) is of degree ml in u, nl in tI. Any given value tI, u has, in general, ml possible values (counting the multiplicities). That is, there are ml values of s for a given tI. Therefore V is of degree ml. In a similar

127

Factorization of Meromorphic Functions

manner, we can verify that U is of degree nl. Since I(z) and g(z) are both entire, U and V can have one pole at finite value ~. Moreover, to any value z, s(z) "I ~o, we may assume without loss of generality, that ~o = 0 (otherwise replace ~ by ~ - ~o). It follows from the above analysis on U and V that for some non-negative integers s($ nd and t($ md ml-t

nl-8

U(~) =

L

ak~k

,

V(d

It follows that Pm

Pm

0

U

0

0

=

L

bk~k

.

(3.26)

k=-t

k=-.

I(z) = Pn 0 g(z) with (3.25) yields

s(z) = Pn 0 V

0

s(z),

Vz E CD (the complex plane) .

Since s(z) assumes an infinite number of distinct values, the above equation leads to PmoU(~)=PnoV(d

V~ECD.

From this result and (3.26), for sufficiently large values of ~, we can derive the following asymptotic relations:

Pm

0

U(d ,.... dl~(nl-.)m,

Pn 0 V(d ,.... d2~(ml-t)n

where d 1 and d 2 are two nonzero constants. Thus (nl - s)m

=

(ml - t)n .

Recalling that (m, n) = 1,0$ nl - s $ n,O $ ml - t $ m we can conclude that nl - s = nl and ml - t = m. Hence either nl = n, s = 0 and ml = m, t = 0 or nl = s, ml = t. This shows that either U and V are polynomials of degree nand m respectively or they are polynomials in 1/ ~. We then can make the substitution W(z) = (noting that s(z) "I o Vz E C), and proceed with the discussion on the polynomials, Udw) = U(~) and VdW) = V(~). We shall arrive at the same conclusion: nl = n and ml = n. The lemma is proven.

./z)

J.F. Ritt made a further study on the equation:

(3.27) where Pm, Pn , Un and Vm are polynomials of degree m, n, n, m respectively; (m, n) = 1. He obtained the following result. ["Prime and composite polynomials," Trans. Amer. Math. Soc. 23 (1922) 51-66]. Lemma 3.1. The Eq. (3.27) can hold only in the following cases:

128

Fix-points and Factorization of Merom orphic Functions

(A) There exists linear polynomials >'1, >'2, >'3 and >'4 such that

>'1 0 Pm 0 >'2(U) = Tm , >'10 Pn 0 >'3 = Tn, >, 2 1 0 Un 0 >'4 = Tn, >.;10 Vm 0 >'4 = T m , >'10 Pm(Un ) 0 >'4 = >'1 0 Pn(Vm ) 0 >'4 = Tnm , where the Tk denotes kth Chebyshev polynomial. (B) Suppose m > n. There exists linear polynomials >'1, >'2, >'3, >'4 and a polynomial h of degree less than such that

min

>'10 Pm 0 >'2(U) = u 2h(u)n (r + ndeg h >'2 1 Un 0 >'4(S) = sn , >'10 Pn0 >'3(U) = un , >.;10 Vm 0 >'4(S) = Sr h(sn) ,

>'10 Pm (Un) 0 >'4(S)

= m)

,

= >'1 0 Pn(Vm) 0 >'4(S) = [Sr h(sn)t

.

Proof. Without loss of generality we may assume, 1 < n1 < q < n2 < n3 < ... ; nj+1

~

qnj, i

=

1,2, ...

(3.28)

Applying Lemma 3.5 with m = njU ~ 2), n = q, we get polynomials U (of degree q), V (of degree m), and entire function sm(z) such that:

(3.29) Now according to Lemma 3.7, the polynomials Pm, U, Pq , and V must satisfy one of the cases, (A) or (B). We shall show that if case (B) holds for a certain pair (m = nj, n = q), then (B) will hold for any other pairs (nk' q). Alternatively suppose that we consider case (A) for a certain pair (nk' q). There exists linear polynomials p and (1 such that, po Pq 0 (1(u)

= Tq(u) ,

where Tq is qth Chebyshev polynomial. On the other hand, case (B) holds for m and n = q < m. Thus there are linear polynomials J.L, >. such that

Hence

Factorization of Meromorphic Functions

129

or

A(Bu + C}q

+D =

Tq(u} .

Note that Tq does not assume any value of multiplicity ~ 3. The above equation leads to a contradiction, since q ~ 3. Therefore, Theorem 3.12, will be a consequence of the results of the following two results, Theorems 3.13 and 3.14.

Theorem S.lS. IT there exists an infinite sequence M = {mk} and integer q ~ 3 satisfying (q, mk) = 1 (for k = 1,2, ... ) such that (3.29) and case (A) of Lemma 3.7 hold, then F must have the form: F(z} = acos VH(z} + b; where a(¥= o} and b are constants, and H(z} is an entire function. Theorem S.14. IT there exists a sequence of values of m( = {mk}} and n = q satisfying the condition of Eq. (3.28) such that Eq. (3.29) and case (B) in Lemma 3.7 hold, then F must have the form:

F(z}

= aeH(z)

+b

where a(¥= o} and b are constants and H(z} is an entire function.

Proof of Theorem S.lS. By case (A), corresponding to each m EM, there exists linear polynomials .Am and 11m such that

IT ~m and £1m are the linear polynomials that correspond to we have \ ,-1 Am 0 Am 0

T.q

--1 Ollm 0 11m

m EM.

Then

= T.q.

Recalling that for q ~ 3,Tq(u} cannot be expressed as A(u - a}q + B, from this and Lemma 3.5, we can conclude that there can only be a finite number of pairs (.Am 0 ~;;.1, £1;;.1 Ollm ). Keeping m fixed and replacing M by an infinite subsequence N, if necessary; we may assume that .Am does not depend on the choice of m. It follows from Eq. (3.29) and (A) that there exists a linear polynomial .A such that

where Sm(Z} is a composition of the function sm(z} in Eq. (3.29) and a linear polynomial.

130

Fix-points and Factorization of Meromorphic Functions

Setting (3.30) we have ). 0

F(z) = cos(qmtP(z)) .

(3.31)

The expression tP in Eq. (3.30) is a multivalued function but in a disk D (in the z-plane), that contains no roots of srn(z) = ±1, we can define tP without ambiguity as a specific branch of cos- 1 srn(z). IT tPo is one of the branches of cos- 1 srn(z), then any other branch tP(z) is determined by the formula: tP(z) = ±tPo(z) (mod2n-), Vz ED. Let n be a member in N other than m. We can obtain functions sn(z) and ,p(z) having properties similar to that of srn(z) and tP(z) respectively such that

). 0

Sn(z) = cos,p(z) ,

(3.32)

F(z) = cos(qn,p(z)) .

(3.33)

Similarly, for the multivalued function ,p, we can define in a disk (that incidentally can be chosen to be identical to D) a specific branch ,po such that ,p(z) = ±,po(z) (mod 21r), Vz ED. (3.34) From Eqs. (3.31) and (3.33) we have

qn,po(z) = ±qmtPo(z) (mod 21r) . We may suppose (if necessary, by changing ,po into -,po)

qn,po(z) Hence,

=

qmtPo(z) + 2k1r .

m 2kn,po(z) = -tPo(z) + .

n

We set ~o(z) -

= ,po(z) + 2tn-, m n

,po(z) = -tPo(z) +

Cj

(3.35)

qn

where t is suitable integer, such that C

is a constant satisfying - n- <

C

:$ n- .

131

Factorization of Meromorphic Functions

Both
= o
-

,pr(z)

0 = ±1, where 1 is an integer.

m

m

2ml1l"

= -
(3.36)

and also ,pr(z)

-

m

= '7,p(z) + 2k1r = '7-
'7

=

±1 .

Comparing the above two expressions and noting that k depends on nand m and r, and c on m and n only. We see that 6' = '7. Therefore

r, 1 on

c+ 2lm1l" /n = '7C + 2k1r . If '7 = 1, this leads to 1~ = k. Since n must be sufficiently large, we must conclude that k and hence 1 are both zero. If '7 = -1. We have 2k1l" = 2c + 2lm1l"/n. By assuming -11" < C ~ 11" and for large n, we get k = 0 or -1. Since c is independent of the path r, when '7 = -1, k is also independent of r. Consequently the relationship between ,pr and ~o can be summarized into the following cases. (i) k = O,,pr = ~0('7 = 1) or ,pr = -~o ('7 = -1) (ii) k = 1,,pr = -~o or ,pr = 211" - ~o (iii) k = -1,,pr = ~o or ,pr = 211" - ~o. Thus we can see that ,p2, (,p _11")2 and (,p+1I")2 are single-valued analytic functions of z where sn (z) t= ±1. Now,p is locally defined to be a branch of cos- 1 (sn (z)) and approaches a finite value as z tends to a root of equations sn(z) = ±1. Therefore, all these roots are removable singularities of the above three functions. Under case (i)

is an entire function. Moreover, >. 0 F(z) = cos y'H(z), and F(z) assumes the form in Eq. (3.24). For the other two cases, we have

,p(z) ± 11" = y'H(z)/qn .

132

Fix·points and Factorization of Meromorphic Functions

Therefore, >. 0 F(z) = cos(=t=qmr + ylH(z)) = ± cos ylH(z). Thus F(z) assumes the form in Eq. (3.24). The theorem is thus proved.

Proof of Theorem 3.14. Letting m = nk and n = q, in Eq. (3.29) and relation (B), we see that there exists a linear polynomial and entire function Sk (z) = Aks nk (z) + Bk such that (3.37) where the hk are polynomials with hk(O) ::f 0, and rk + q deg hk = nk. We may assume without loss of generality that hk(O) = 1. By applying the last form in relation B for the indices nl and q, there exists a linear polynomial >'0 such that (3.38) IT or and /3 are roots of >'o(t) = 0 and >'k(t) = 0 respectively, we shall show that or = /3. Assume that Q::f /3, then by Nevanlinna's second fundamental theorem, we have

for a suitable sequence of r values tending to 00. On the other hand, we observe that each root of the equations, F(z) = and F(z) = /3 has multiplicity of at least nl and q, respectively. Hence

Q

-( 1) ::;-N 1 (r'-F 1) ::;-(I+o(I))T(r,F) 1 N r'-F , -

and

N (r, F

Q

nl

-

Q

nl

~ /3) ::; ~N (r, F ~ /3) ::; ~(1 + o(I))T(r, F)

.

Substituting the above two results into Eq. (3.39) and noting that N(r, F) = 0, we get

T(r, F) ::;

(~+ -..!.. q nl

+0(1)) T(r, F) ,

which is impossible. It follows that Q = /3, and hence >'k(t) = c>'o(t). Replacing Sk (t) by 6Sk (t), if necessary, we may assume that c = 1, i.e.,

>'0

=

>'k

(k = 2,3, ... ) .

Factorization of Meromorphic Functions

133

Thus

(3.40) By assuming (n1' q) = 1, we easily see from the above expression that the multiplicity of every root of Fo(z) = 0 is divisible by n1q. Hence, we have

where G is an entire function. From this and Eqs. (3.38) and (3.40), we have

We let a

where J.L( a) denotes the multiplicity of the root anI. Then n,

h(t n ,) =

II II (t a

p;a)l'(a)

;=1

where p is a primitive n1-th root of unity. IT Zl is a zero of s(z) - ria with multiplicity /.I, then ql/.lJ.L(a) Suppose that qf'J.L(a) then /.I ~ 2, and ria is a completely ramified value of s(z). Also r + n 1deg h = q,O < r < q. It follows that every root of s = 0 has a multiplicity ~ 2. Thus if deg h > 0 and h has some zero with a multiplicity not divisible by q, then s has at least three completely ramified values. This is impossible. Hence we see that there are only two possible cases that arise

(i) srh(snl) = sq or (ii) qlJ.L(a) for every zero of h .

(3.41)

However, case (i) implies srh(snl) = sr(H(sn,))q where His a polynomial. It also implies that q = r + n1 deg h = r + n1q deg H. This is impossible for deg H > o. Thus only case (ii), i.e., Eq. (3.41) can hold. Therefore, Eq. (3.37) becomes

Fo(z) = s(zt,q . Letting k

= 2 in

Eq. (3.40), we get

134

Fix-points and Factorization of Meromorphic Functions

and we may also assume

Substituting 8 by 82 and G by 8, and exchanging the positions of q, we can, using the reasoning above, obtain

where K2 conclude

IS

a polynomial Since

r2

> 0 and n2

~ nlq, (n2' q)

nl

and

1, we

Hence,

By repeating this kind of argument, we are led to

This shows it is impossible for Fo(z) to possess any zeros_ Thus

Fo(z) = >'oF(z) =

eH(z)j

H(z) an entire function.

Since we have shown that F has form (3.23)' Theorem 3.14 is proven. This also concludes the proof of Theorem 3.12. 3.5. FACTORIZATION OF ELLIPTIC FUNCTIONS In the previous section 3.4 we discussed the possible factors of the periodic cosine function and realized that they are quite restricted. Now we shall study the possible factors of an elliptic function h(z). More specifically, we would like to find when h(z) = J(g(z)) what forms and properties, J (left factor) and 9 (right factor) may possess.

Theorem 3.15. factor.

No elliptic function h(z) may have a periodic left

Proof. First we recall that the order and lower order of an elliptic function h are both equal to 2. This is because any elliptic function can be expressed as a rational function of a sigma function and its derivatives.

135

Factorization of Meromorphic Functions

Now the sigma function (and hence its derivatives) has order 2, so that the order of h (= p(h)) ~ 2. On the other hand, to any value "a",

T(r_1 »N(r_1»~ ,h- a

-

,h- a

log r

for some constant c. This leads to the conclusion that p(h) 2: 2. Therefore we arrive at p(h) = 2. We now suppose that h is not prime and has the factorization h = fog. IT f and 9 are both transcendental entire functions then, by P6lya's theorem, p(j) = 0, and f cannot be a periodic function. Also if f is periodic, say period 1, then 9 must be a polynomial. Consider a point set S = {zlz is a root of one of the equations g(z) = m + c, m = 0, ±1, ±2, ... j c is a constant}. We shall show that S has one finite limit point. Let g(z) = AkZk+ ... +A1Z+Ao and Zm,Zm+j be the roots satisfying g(z) = m + c, g(z) = m + i + c respectively. We have

Ig(zm) - g(zm+j)1 = IZm+j - zmIIAkllz~-~/j + ZmZ~+2j + Z~-l + Pk- 2(Zm, zm+j) I = i ,

+ ... (3.42)

where Pk - 2 is a polynomial with variables Zm and zm+j of degree k - 2 at most. We can easily see that for each m, there always exists integers il, J2 2: m such that lil - J21 < 4k2 and I arg Zl - arg z21 ~ ~. We can thus derive

which approaches 00 as m --+ 00. This would imply from Eq. (3.42) that IZm+j - Zm I tends to as m --+ 00. Therefore, S must have a limit point. Now we have h = foP where P is a polynomial. Assume that h has periods Tl and T2 and recall that f has period 1, then the following identity holds for any integers m, nl and n2.

°

(3.43) For any fixed Zo we set

From the above analysis, we see that S has a finite limit point. Thus Eq. (3.43) will yield a conclusion that f is a constant, which is a contradiction. The theorem is thus proven.

Fix-points and Factorization of Meromorphic Functions

136

With regard to the right factors of elliptic functions we have the following result.

Theorem 3.16. Let h(z) be an elliptic function of valence 2. If h = fog for some transcendental entire function f and an entire function 9 that is not a linear polynomial, then the right factor 9 must be either (i) a polynomial of degree 2 or (ii) of the form A cos(z + r) + B, where A, B, and r are constants. Proof. By assuming that h is an elliptic function of valence 2, it means that h satisfies the following differential equation:

(h')2

=

P(h) ,

where P is a polynomial. Thus we have

(g' /,(g))2 = P(I(g)) . If follows that '2

9

=

P(I(g)) (I' (g))2

=

F( ) g,

where F(~) = P(I(d)/ f'(d 2, a meromorphic function. According to a theorem of Clunie's, we conclude that F(~) must be a ra. If . 0 h . 1· T(r,F(g)) ·11 1· -1· T(r,F(g)) tiona unction. t erwlse 1m T( ) = 00 WI resu t In 1m T( 2)

=

r-+oo

r,g

r-+oo

r,g'

00.

Therefore, we have c is a constant

=1=

0 .

(3.44)

Note that since g' has no poles, the denominator in the above expression never vanishes. Moreover, by examining the multiplicities of the roots of g(z) = ai, we can verify easily that 9 has at most two complete ramified values, say al and a2. Then Eq. (3.44) becomes

From this we conclude that if 9 is a polynomial then deg 9 = 2, and nl = = 1 if 9 is a transcendental entire function. In the latter we have

n2

137

Factorization of Meromorphic Functions

gl2 = ddg(z) + d2)2 + d3 , where d1 ,d2, and d3 are constants. It follows that g(z) has the form Acos(cz + r) + B. This also completes the proof. Remark. It is not difficult to exhibit some elliptic functions have transcendental right factors. Sn(z), Cn(z) and dn(z) are such functions. •

Sn(2kz/7r) = csmz

II 00

(

n=l

1 _ 2q2n cos 2z + q4n ) 2 1 4 2 1 - 2q n- cos 2z + q n-

We easily see that Sn(2kz/7r) = J(sinz)' where

J(d = c~

II 00

(

1- 2q2n(1- 2~2) 1- 2q2n-l(l- 2~2)

+ q4n ) + q4n-2 '

a meromorphic function.

Earlier in this chapter, we showed that if J is a transcendental entire function and P an arbitrary polynomial of degree ~ 3, then J 0 P cannot be periodic. We conclude this section by proving the following result.

Theorem 3.17. Let J(z) be a non-constant meromorphic function and P(z) be a polynomial of degree n. Then F(z) = J oP(z) cannot be periodic unless n = 1,2,3,4 and 6. Proof. It is, of course, possible for n = 1 and 2. Therefore we shall only deal with the cases when n ~ 3. Suppose that F is a periodic function and by changing variables if necessary, we may assume without loss of generality that F has a period of 1. Moreover, we may assume P(z) has the form

P( Z ) = aoz n + an-tZ n-t + an-t-lZ n - t - l + ... + ... where t is an integer equation

~

2. It is clear that for any given z and the following

P(~)

= P(z + m)j

JzJ > ro ,

(3.45)

always has a root. Furthermore, for sufficiently large m, we have ~

= ,,(z + m) + 0(1) (m -+ 00) ,

(3.46)

where" = e21ri / n • We observe that, for sufficiently large m, any integer m ' (from the above we have k + m'l) will be greater than ro as in Eq. (3.45). Since F has period 1, F(~)

=

J(P(~)) =

J(P(z

+ m)) = f(P(z))

=

F(z) .

Fix-points and Factorization of Meromorphic Functions

138

On the other hand, to the ~ in Eq. (3.46), the equation P(~') = P(~

has a root ~' satisfying ~' m -+ 00. Thus F(~'

=

11(~

+ m')

+ m') + 0(1) = I1 2 z + 11 2 m + 11m' + 0(1)

as

+ m) = F(~') = F(d = F(z) .

Consequently to a given point z, the equation

F(w) = F(z)

(3.47)

always has a root w satisfying

for any given integer m' and Iml sufficiently large. Since 11 = e21ri / n , we have

Suppose that 2cos

2: is a irrational number.

To any given real number

f3, we can choose m and m' such that the right-hand side of the above expression can be made arbitrarily close to f3. Thus F(I1 2 z

+ 11f3) = F(z) (-00 < f3 < 00) .

However, z is an arbitrarily given number, hence, from the above equation, F must be a constant function. This creates a contradiction. Suppose that cos = a is a rational number. Then the nth root of unity 11 will satisfy the equation

2:

In the meantime, 11 must satisfy an irredicuble equation g(I1) = 0, where the degree of g is cp( n)( cp( n) denotes the Euler function of n). Therefore, 112 - 2al1 + 1 must be divisible by g(I1). Hence cp(n) = 1 or 2_ However,

cp( n) = n

II (1 qln

~) ~ II (q gin

1) .

Factorization of Meromorphic Functions

139

It follows that if ip(n) ~ 2, then n can only have 2 and 3 as its factors. This results in n = 3,4 or 6. The theorem is thus proved.

Discussion. Illustrate by examples, that for n = 3,4 or 6 there exists a meromorphic function In and polynomial Pn(z) of degree n such that

In(Pn(z)) is an elliptic function. 3.6. FUNCTIONAL EQUATIONS OF CERTAIN MEROMOPRHIC FUNCTIONS Factorization theory can be included in the theory of functional equations. The factorization of F(z) = I(g(z)) can be viewed as the finding of functions F, I and 9 that will satisfy the expression just mentioned. Various forms of functional equations have been derived in the course of studying problems relating physics, practical or theoretical mathematics. For example, a well-known problem is Cauchy's functional equation: I(x + y) = I(x) + I(Y). In general, it is difficult to obtain a concrete solution to a functional equation. Many have obtained results and focused their research on the necessary and sufficient conditions for the existence of solutions or certain special properties of the solutions. Here we shall introduce certain simple forms of the functional equations of meromorphic functions to show the existence as well as the growth properties of the solutions. We shall first discuss the following type of equation:

I(g(z)) = h(z)/(z) , where we restrict following results:

I, 9

(3.48)

and h to entire functions. It is easy to derive the

Theorem 3.18. Let I(z), g(z) and h(z) be entire functions and satisfy the Eq. (3.48). Suppose that h(z) is a polynomial and 9 is not a linear polynomial. Then I must be a polynomial. Theorem 3.19. Let I(z), g(z) and h(z) be entire functions and satisfy Eq. (3.48). Suppose that both I and h are non-constant polynomials, then g must be be a polynomial. Theorem 3.20. Let g(z) == z2 and I, 9 be non-constant entire functions. Assume that h has only a finite number of zeros. If Eq. (3.48) holds for such I, g, and h, then I must have only a finite number of zeros as well. Most of the results introduced here were obtained by R. Goldstein who also extended the previous discussion by considering meromorphic solutions

140

Fix-points and Factorization of Meromorphic Functions

of the following type of equation:

I(g(z)) = h(z)/(z) + H(z) _

(3.49)

Theorem 3.21. Let I, g, hand H be meromorphic functions satisfying Eq. (3.49). Suppose that I, g are non-constant functions and g is always a transcendental entire function unless I is a rational function_ Also suppose that there exists a positive constant k such that for r> ro (a constant),

T(r, h) T(r, H)

~ ~

kT(r, f) , kT(r, f) .

(3.50)

Then g must be a rational function of, say, order m_ Furthermore, if transcendental, then 1 ~ m ~ k + 1, and when m > 1, I must satisfy

where e is any given positive number and a

I

is

= log(2k + l)/logm.

Proof. From Eqs. (3.49) and (3.50), we have

T(r,J(g))

T(r,f) + T(r,h) + T(r, H) + 0(1) < (2k + l)T(r, f) + 0(1) . ~

But Clunie proved that if

I

(3.51)

and g are transcendental, then lim T(r,l(g)) =

r-+oo

T(r, f)

00

(3.52)

which will contradict with Eq. (3.51)_ Hence I and g cannot both be transcendental. It is clear that if I is rational function and g is transcendental, then from Eq_ (3_50) we conclude both hand H are rational functions. As a result Eq_ (3.49) will not hold. For this equation to hold, I must be transcendental and g must be a polynomial. Before proceeding further we prove the following result.

Lemma 3.S. Let !/I(r) be a positive and continuous function of r satisfying, for some m > 1, (3_53)

141

Factorization of Meromorphic Functions

where p., A(A > 1) are two positive constants. Then

¢(r) = O((logr)a)j

Q

= logA/logm.

Proof. We put in Eq. (3.53) log p. 1-m'

¢(r) = 4>(t)

A4>(t),

(t

t = logr - - that yields

4>(mt) We choose becomes

Q

~

~

to) .

such that m a = A and put ¢(t) = 4>(t)/ta. Then Eq. (3.53)

cp(mt)

~

cp(t),

(t

~

to) .

Now, cp(t) is also a positive and continuous function for t > 0 and the above inequality ensures that it is bounded above by some number B for sufficiently large values of t. Thus

¢(r) = 4>(t)

~

Bt a = B (IOgr _ IOgP.)a

~

Bdlogr)aj

1-m

where Bl is a suitable constant.

Now we continue the proof of Theorem 3.21. Let ( ) =amz m +am-lz m-l + ... +ao gz

(am

i- 0),

m ~ 2,

and p. = laml- 6 (0 < 6 < laml). Since Ig(z)1 ,..., lamlrm for sufficiently large values r, we have for any value "a".

n(r, a, f(g))

~

mn(p.rm, a, f),

(r

~

ro) .

By integrating, we get

N(r, a, f(g)) - N(ro, a, f(g))

~

j

r

mn(p.tm a f) " dt + 0(1) log r . r

ro

We put s = p.tm and obtain

j

r

ro

mn(p.tm, a, f) dt = (Wm n(s, a, f) ds t

JI'r'(J'

S

= N(p.rm, a, f) - N(p.rO' , a, f)

+ O(log r) .

Fix-points and Factorization of Meromorphic Functions

142

By combining the above two inequalities, we have

N(J.l.r m , a, I) +O(logr)

~

N(r,a,/(g)),

(r ~ no).

(3.54)

But, it is well known that for a suitable value a,

N(r, a, I) ,... T(r, I) and log r = o(T(r, (3.51) we obtain

I))

for a transcendental function

T(J.l.r m , I)

~

(2k + 1 + e)T(r, I),

I.

By Eqs. (3.54) and

r ~ rl .

Applying Lemma 3.8, the required result follows. By a similar argument we can obtain the following result:

Theorem 3.22. Let I, 9 and h be non-constant meromorphic functions satisfying Eq. (3.48). Suppose that 9 is always a transcendental function unless I is rational. If there exists a positive constant k such that

T(r, h)

~

kT(r, I),

(r

~

ro)

then 9 must be a rational function of order m. Furthermore, if m > 1 and e is any given positive number, then unless I is rational, m ~ k + 1, and

T(r, f) where

f3

=

= O(log r)H..

as r

-+ 00 ,

log(k+ l)/logm, and

T(r,h) --- > m - 1 - e T(r, I)

(r

~

rt} .

(3.55)

In the following we shall investigate Eq. (3.48), in which the zeros or poles of h have been restricted. We shall call the value a a Fatou exceptional value of 9 of multiplicity m if g(z) == a + (z - a)meG(z); where G(z) is an entire function.

Theorem 3.23. Let I and h be meromorphic functions, and 9 be a nonlinear entire function satisfying Eq. (3.48). Suppose that h has no poles (zeros)' then I has at most one pole (zero) at z = 0:, and a is a Fatou exceptional value of 9 of multiplicity 1. Proof. We give the proof for the case where h(z) has no poles. The case where h has no zeros can be proved similarly. Using the assumption,

Factorization of Meromorphic Functions

143

we easily see that if z = a is a pole of J(z) and g(b) = a, then z = b must be a pole of J(z). Repeating this argument yields if z = a is a pole of J and if for some n, gn(z) (the nth iterate of g) = a, then z = b is also a pole of J(z). We now need a result of Fatou's as follows.

Lemma 3.9. Let g(z) be a nonlinear entire function. Then there exists a nonempty perfect set T(= T(g)) of complex numbers with the property that to any Zo E T and an arbitrary number w (with one possible exceptional value) there corresponds a sequence of positive integers {nj} U = 1,2, ... ) and a sequence of complex numbers {Zj} U = 1, 2, ... ) such that limzj

= Zo

and

gnj (Zj) = w,

(j = 1,2 ... ) .

(3.56)

This result combined with the conclusion at the beginning of the proof, implies that J has at most one pole, at z = a. Furthermore, z = a must be a Fatou exceptional value of g(z). Hence we have (3.57) where G(z) is an entire function and m is a non-negative integer to be determined. We express J(z) as

J(z) = F(z - a) , (z - a)n

(3.58)

where F is an entire function with F(O) =I=- 0 and n is a positive integer. We will only treat the case where m ~ 1 (a similar argument applies if m = 0). Then Eqs. (3.48), (3.57), and (3.58) yields

F{(z - a)meG(z)} _ h(z) F(z - a) (z - a)nmenG(z) (z - a)n Consideration of the order of the pole at z = a leads to either n = 0 or m = 1. But when n = 0, J becomes an entire function which contradicts

144

Fix-points and Factonzation of Meromorphic Functions

the hypothesis that z = a is a pole of I. Therefore m = 1 and I and g are given by Eqs. (3.58), (3.57) respectively. Theorem 3.23 is thus proven.

Corollary 3.1. Let I, g and h be as in Theorem 3.23. If g(z) and h(z) are nonlinear polynomials, then I(z) has no poles (zeros). Theorem 3.24. Let g(z) be a polynomial of degree m ~ 2, and let I, h be meromorphic functions satisfying the equation I(g(z)) = h(z)/(z). Suppose that I is of finite order and 6(0, h) = 6(00, h) = 1. Then 6(0, f) = 6(00, f) = 1 and Ph = mPI; where Ph and PI are the orders of I, h respectively and must be positive integers.

Proof. It is well known under the hypotheses that P/(g)

= mPI

.

(3.59)

By 6(0, h) = 6(00, h) = 1, h is of regular growth and is of positive integer or infinite order. Now

h( ) = I(g(z)) z I(z) , hence,

Thus

> 0 and (3.60) Ph ~ mPI < 00 . On the other hand, from the equation I(g) = hi and Eq. (3.59) we have PI

(3.61) Since PI < 00, we deduce from the above inequality that

(3.62) Thus, by Eqs. (3.60) and (3.62), Ph

=

mp I

=

PI(g)

< 00 .

From this and the fact that h is of regular growth we have

(3.63)

145

Factorization of Meromorphic Functions

whence lim T(r, h) T(r, I)

= 00 .

(3.64)

"-+00

From the equation I(g)

= hI,

we have

T(r, log) :5 T(r, h) + T(r, I) . Hence from Eq. (3.64) we get

T(r, log) < 1 + 0(1) T(r, h)

-

(as r _ 00) .

Conversely, from h = I(g)/ I and Eq. (3.64) we deduce

T(r,/(g)) > 1 + 0(1) . T(r,h) Thus, we have

T(r, I(g)) '" T(r, h) . Now n(r, 0, I(g)) :5 n(r, 0, h)

(3.65)

+ n(r, 0, I), so

N(r, 0, I(g)) :5 N(r, 0, h) + N(r, 0, I) + O(log r) :5 N(r, 0, h) + T(r, I) + O(log r) . Using Eq. (3.64) and the assumption 8(0, h)

=

1, we have

-1· N(r,O,/(g)) < -1· N(r,O,h) + (1) = (1) 1m T(r,h) - ,.~~ T(r,h) 0 o.

(3.66)

Therefore, by Eqs. (3.66) and (3.65) we have lim N(r,O,/(g)) "-+00

T(r,/(g))

=

lim N(r,O,/(g)) "-+00

= 0(1)

.

T(r, h)

This shows that 8(0, I(g)) = 1. Similarly we can prove that 8(00, I(g)) = 1. Now we prove that PI must be a positive integer. If log is a meromorphic function of finite order, and g is a nonlinear polynomial satisfying I(g) = hI, then by 8(0, I(g)) = 8(00, I(g)) = 1, we also have 8(0, I) = 8(00, I) = 1. (For the proof, we refer the reader to Goldstein's paper "Some results on factorization of meromorphic functions", J. London Math. Soc. (2).( (1971) 357-364.

146

Fix-points and Factorization of Meromorphic Functions

Hence, PI must be a positive integer. This also completes the proof of the theorem.

Discussion. (i) H g(z) is allowed to be meromorphic and h has no poles, what conclusions will result? (ii) Does the condition 6(0, f(g)) = 6(00, f(g)) = 1 always lead to 6(0, f) = 6(00, j) = 1? Theorem 3.25. Let f be a non-constant meromorphic function, g be an entire function, and q(z) be a polynomial of degree k (;::: 1) satisfying the following equation

f(g) = q(l) . Then q(z) must be a polynomial of degree m ~ k. Furthermore, if m > 1, then T(r, j) = O(l)(log r)a, where ex

=

(log k/ log m)

+ e (e is any given positive number) _

We omit the proof, this being analogous to the proof of Theorem 3.21. 3_7_ UNIQUENESS OF FACTORIZATION For simplicity we shall only discuss entire functions and their entire factors_ We state that two factorizations (of entire factors) fdh(··· (In)) ... ) and gdg2(" . (gn)) ... ) are equivalent if there exists linear transformations .Al, _. - ,.An-l such that

An entire function F is called uniquely factorizable, if all its factorizations of nonlinear prime entire factors are equivalent to each other_ Ritt obtained a complete answer to the uniqueness factorization problem for polynomials (see the appendix). The result essentially states that, besides the following three non-equivalent cases, for pairs of consecutive factors fdh) and gl (g2), the two factorizations of a polynomial F(z) will be equivalent. The exceptions are: (i) h(z) = zk, h(z) = zl and gl (z) = zl, g2(Z) = zk (ii) h(z) = zk[h(z)]l, h(z) = zl, and gdz) = zl, g2(Z) = zkh(zl), and (iii) h(z) = Pk(Z), h(k) = Pl(z), and gdz) = PI(Z)' g2(Z) = Pk(Z), where Pk(Z) is the kth polynomial satisfying cos kz = Pk(cos z)_ Case (ii) may also arise in the factorization of a transcendental entire function. H, for example, we let F(z) be zP exp zP (p is a prime number)'

Factorization of Meromorphic Functions

147

then F has two factorizations that are not equivalent: F(z) = zP 0 (ze ZP Ip) and F(z) = (ze Z) 0 zp. However, it is not difficult to show that both F(z) = zPePZ(= zP 0 ze Z) and F(z) = (ze Z) 0 (ze Z) are uniquely factorizable. Moreover, the latter is almost the simplest function one can show in demonstrating the uniqueness factorization of transcendental entire factors. As a generalization, H. Urabe obtained the following result in his dissertation.

Theorem 3.26. Let F(z) = (ze Z) 0 (h(z)e Z), where h(z) is a nonconstant entire function of order less than one, and has at least one simple zero. Then F is uniquely factorizable. Proof. (sketch) Let F(z) = (ze Z) 0 (h(z)e Z) = I(g(z)); I, g being two nonlinear entire functions. By virtue of the assumption that h(z) has at least one simple zero and the Tumura-Clunie Theorem we conclude that I cannot be a polynomial. According to a result of Edrei-Fuchs' that if I and g are two transcendental entire functions with the exponent of convergence of the zeros of I being positive, then the zeros of I(g) have an exponent of convergence equal to infinity. Therefore, we need only consider three cases: (a) I(z) = hdz)eP(z), hI nonlinear entire function with p(h) = 0, p(z) is a non-constant polynomial, and g(z) is a transcendental entire function with p(g) < 1; (b) I(z) = zeP(z), g(z) = h(z)eq(z), where p, and q are non-constant entire functions; and (c) I(z) = hdz)eP(z), where g(z) is a polynomial of degree ~ 2, and hI and p are non-constant entire functions satisfying p(hd < de~g (hence p(hd(g) < 1). In case (a), from F = I(g), we obtain hl(g(z)) = h(z)ed(z) and p(g(z)) = z - d(z) + h(z)e Z, where d(z) is an entire function with p(d) < 1 (Polya's Theorem, Corollary A.1, Appendix). By Theorem 4.2, it follows that p(z) must be a polynomial. Thus p(g) = 1, which is a contradiction. In case (b), we obtain a functional equation q(z) +p(h(z)eq(z)) = z+h(z)e z . It is easily verified from this relationship that q(z) must be linear and the uniqueness of factorization follows. In case (c), we have the relations hdg(z)) = h(z) and p(g(z)) = z + h(z)e z . Using these equations and an argument similar to the proof of case (b) we easily arrive at the uniqueness of factorization of F. Urabe also obtained the following more general result. Theorem 3.27. Let F(z) = (z + h(e Z)) 0 (z + q(e Z)), where h(z) is a non-constant entire function with the order p(h(e Z)) < 00 and q(z) is a

148

Fix-points and Factorization of Meromorphic Functions

non-constant polynomial. Then F(z) is uniquely factorizable. We note that eZ and cos z are pseudo-prime and have an infinite number of different factorizations. There exist some transcendental entire functions that are not pseudoprime but have an infinite number of equivalent factorizations. For example, if we let F(z) = z - sin z + sin (sin z - z). Then F(z) = 101 = gog where I(z) = z - sin z and g(z) = sin z - z + 2k1r(k integer =I- 0). The following questions are therefore interesting.

Question 1. Do there exist two nonequivalent factorizations 11 0 12 gl 0 g2, where 11, 12, gl, g2 are prime transcendental entire functions?

=

Question 2. (Gross) Do there exist prime nonlinear entire functions h, 12,··· ,1m and gl, ... ,gn with n =I- m such that

h 012

0 ••• 0

1m == gl

0

g2

0 ••• 0

gn?

4 FIX-POINTS AND THEORY OF FACTORIZATION

4.1. THE RELATIONSHIP BETWEEN THE FIX-POINTS AND THEORY OF FACTORIZATION We have shown in the previous chapter that eZ + z is prime. Gross conjectured that functions F(z) of the form

F(z) = Q(z)e<>(z)

+z

,

(4.1)

where Q(z) is a polynomial and a(z) is a non-constant entire function, must be prime. To date, the conjecture has not been answered. * However, some partial results have been obtained. Most of these were stated in terms of fixpoints. Recall that at the beginning of Chapter 3 we proved that if P(z) is a nonlinear polynomial and 1 is a transcendental entire function, then P(I(z)) has an infinite number of fix-points (Theorem 3.2). We now prove the following lemma.

Lemma 4.1. Let

1

and 9 be two non-constant entire functions. If

I(g) has only a finite number of fix-points then g(l) also has only a finite number of fix-points.

Proof. Let Zo be a fix-point of I(g). That is, if I(g(zo)) = Zo, then g(l(g(zo)) = g(zo). Thus g(zo) is a fix-point of g(l). Moreover, if Zl and Z2 *Recently (1988) W. Bergweiler confinued this (and hence conjecture 1 in next page) in his preprint entitled "Proof of a conjecture of Gross concerning fix-points" by utilizing Wiman- Valiron type of argument.

149

Fix-points and Factorization of Meromorphic Functions

150

are two distinct fix-points of I(g), then g(Zl) and g(Z2) will be two distinct fix-points of g(f)- IT g(zd = g(Z2), then ZI = f(g(zd) = l(g(Z2)) = Z2This proves the lemma_ From this lemma and Theorem 3_2 we conclude that I(P(z)) has an infinite number of fix-points for any transcendental entire function I and nonlinear polynomial P(z)_ We also know that if I, g are two transcendental entire functions, then either g or I(g) must have infinite number of fix-points. It is easy to see then that Gross' conjecture is equivalent to the following conjecture:

Conjecture 1. IT I and g are two nonlinear entire functions, with at least one of them being transcendental, then I(g) must have an infinite number of fix-points. 4.2. CONJECTURE 1 WITH p(f(g)) <

00

Conjecture 1 with the additional hypothesis that p(f(g)) < 00, has been validated by Goldstein, Yang and Gross, and Prokopovich, each using different approach and frames. We now exhibit Prokopovich's statement and its proof as follow:

Theorem 4.1. Let (4.2) where Ql, Q2 and P are polynomials with QI (z) 1- constant. P(z) 1constant, and Q2(Z) 1- O. Then F(z) is prime unless there exist polynomials ql, q2, q3, T, U, V and nonlinear polynomial g(z) such that F = I(g) with and

Qt{z) = qt{g(z)), Q2(Z) = q2(g(Z)), p(z) = q3(g(Z)) , I(z) = T(z) + U(z)ev(z) .

Before proving the theorem, we first quote the following useful fact.

Lemma 4.2. Let P(z) and Q(z) be polynomials of degrees p, q respectively. Suppose that q ~ 2, qlp, and that P(z) is not a function of Q(z). Then P(z) can be expressed as

P(z) = Pt{Q(z)) + P2(Z) ,

(4.3)

where PI and P2 are polynomials and qfdeg P2.

Proof of Theorem 4.1. First of all, it is not difficult to show that F is not a periodic function. Hence it suffices to show that F is E-prime.

151

Fix-points and Theory of Factorization

Suppose that

(4.4)

F(z) = f(g(z)) , where f and g are nonlinear entire functions. Differentiating Eq. 2.1, we obtain

F(z) = j'(g(z))g'(z) = Q~(z)

+ S(z)eP(z), S(z) == Q~(z) + Q2(Z)P'(z) .

(4.5) We now see that the function
F'(zn) = Q~(zn) + S(Zn)eP(z,,) = 0,

n

=

1,2, ....

Thus

It follows that

F(zn) = Qt{zn) =

+ Q2(Zn)e P(z,,)

[S(zn)Qt{zn) -

Q2(Zn)Q~ (zn)]jS(zn)

.

Also from Eq. (4.5), we see that deg S ~ deg Q2, and hence t = deg(SQl Q2Q'1) > degS. This shows that in {zn} there exist no more than 2t distinct points

Znj'

j

=

1,2, ... ,2t such that (4.6)

First we treat the case that g is a transcendental entire function. By a theorem of P6lya we know that f is of zero order. Suppose there exists a point S-o that is a zero of f' but not a Picard exceptional value of g. This implies that there exists an infinite sequence of points {zj} such that g(zj) = S-o. But then F'(zj) = f'(S-o)g'(zj) = 0, j = 1,2, .... This shows that {zj} C {zn}. Furthermore, we have F(zj) = f(g(zj)) = f(S-o), j = 1,2, ... , which contradicts with the result in Eq. (4.6). Therefore, we conclude that f' can have only one zero, say S-o. Moreover, S-o must be a Picard exceptional value of g. Since f' is of zero order, it follows that f' must be a polynomial with the form

Fix-points and Factorization of Meromorphic Functions

152

and

g(z) = s"o + a(z)eb(z) , where a(z) and b(z) are polynomials with b(z) ~ constant. Thus,

F'(z) = Aam(z)[a'(z) + a(z)b'(z)]e(m+1)b(z) . This implies that F'(z) has only a finite number of zeros, a contradiction. Now what left is the case where g(z) is a polynomial of degree n ~ 2. Assume that deg Ql = a, deg P = f3 and note that now f'(s") must have an infinite number of zeros s"l, s"2, .... Let zY) ,k = 0, 1, 2, ... be the roots of the equation g(z) = s"i' It is easy to verify that when 0 ~ k ~ n - 1 and j sufficiently large, we have zY) = (1 + o(1))rie2trik/n, ri --+ 00 U--+ (0) . (4.7) Let j be fixed and zi, z;. be two points in {z;k)} corresponding to k and k = 1 respectively. Then

F'(zi) = F'(zj) = 0,

=

0

S(zi)eP(z;j = -QdZi)

and

S(zj)eP(zj) = -Qdz;.) . By F(zi)

= f(s"i) = F(z;.),

we have

Thus

Qt{zi) - Ql(Z;') = Q2(z;.)e P(zj) - Q2(zi)e P(z;j -

Q2(zi)Q~(zi)

Q2(zi)Q~ (z~.)

S(zi)

S(zj)

(4.8)

Three cases arise that are to be treated separately. Case (i). nf'a. From Eq. (4.7), we have, for sufficiently large j,

Ql(zi) - Qt{z~.) = a(l + o(1)rj(e2tria/n - 1) ,

(4.9)

153

Fix -points and Theory of Factorization

where a is the coefficient of the leading term in Ql(Z), It is also easily verified that

Q2(Zj)Q~ (Zj) _ Q2(zj)QHzj)

= 0(1)

S(z~.)

S(Zj)

~-11 rJ

(4.10)

. -+ ,

J

00 •

Combining Eqs. (4.7) and (4.10) yields:

a(l

+ o(1»)(e 2 .... a / n

-

1)1

= 0(1)

.

This is clearly impossible. Thus case (i) is ruled out. Case (ii). nla and Qdz) cannot be expressed as a polynomial in g. Then by Lemma 4.2, we have

Ql(Z) = T(g(z)) + U(z) , where T and U are polynomials with nfdeg U. We put

G(z)

= F(z)

- T(g(z)) = h(g(z» = U(z)

= J(g(z)) - T(g(z)) + Q2(z)e P (%) ,

= J - T. By applying the argument used in Case (i) to G(z) = h(g(z»), we arrive at a similar contradiction. Thus Case (ii) is ruled out.

where h

Case (iii). nla and Ql(Z) = T(g(z»), where T is a polynomial. From QdZj) = Ql(Zj) = T(S-j) and Eq. (4.8) we have

Q2(Zj )T'(S-j)g' (Zj) _ Q2( zj )T' (S-j )g'( z;.) S(Zj) S(z;.)

(4.11)

Note that T'(s-) 1= 0 and T'(S-) has only a finite number of zeros. It follows from the above equation that there exist an infinite number of i's such that

Q2(Zj)g'(Zj) _ Q2(Z;.)g'(z~.) S(Zj) S(z;.)

For the rest of the proof, we may assume that J. large). It is easily verified that ( (k)g'( (k)

Q2 Zj (k) Zj S(Zj )

_ 1\'(I + 0 (1»

-

n-fj

rj

e

(4.12)

~ Jo

-21r .. kfj/n

( .

(i.e., J. is sufficiently

J -+

00

)

,

(4.13)

154

Fix-points and Factorization of Meromorphic Functions

where). is a constant "I O. IT nfP, then from the above two equations, we have ).(1 + o(1))ri-~(e-21r~/n - 1) = 0 , which is impossible. Finally, if niP but P(z) is not a polynomial g, then P(z) can be expressed as P(z) = V(g(z)) + W(z) where V, W are polynomials with nfdeg W. Then, as before, a similar contradiction will result. Summing up the above discussions, we may conclude that if F(z) = Ql(Z) + Q2(z)e P(z) can be expressed as I(g(z)), where 9 is a polynomial, then Ql(Z) = T(g(z)) and P(z) = V(g(z))j where T and V are polynomials. Moreover,

- T(g(z)) _ U( ( )) Q2 (Z ) -_ I(g(z)) eV(g(z)) 9 z , where U(~) = [f(~) - T(dle-V(r). Since Q2 is a polynomial, it follows that U is also a polynomial. This completes the proof of the theorem. Conjecture 1 with the additional hypothesis that p(f(g)) < in terms of fix-points as follows.

00

is proven

Corollary 4.1. Let I and 9 be two nonlinear entire functions with at least one of them being transcendental. IT p(f(g)) < 00, then I(g) must have an infinite number of fix-points. This is equivalent to saying that Q(z)eP(z) +z is prime for any polynomial Q(z)(t 0) and P(z)(t constant). 4.3. SOME GENERALIZATIONS Theorem 4.2. (Goldstein) Let Pm(z) be a polynomial of degree m(~ 1). Let tPm(z)(t constant) and ,pm(z)(t constant) be entire functions of order < m. Suppose that F(z) = tPm(z) + ,pm(z) exp(Pm(z)) = I(g(z)) for some nonlinear, entire functions I and g. Then g(z) is a polynomial of degree k ~ m. Moreover, if PI denotes the order of I, then kpI = m. Almost concurrently Prokopovich obtained the following result. Theorem 4.S. Let F(z) = tPdz) + tP2(z)e P(z), where P(z) is a nonconstant polynomial and tPl(Z)(t constant), tP2(Z)(t 0) are two entire functions satisfying

T(r, tPi)

=

o(l)T(r, F), (j

=

1,2),

as

r -

00 .

(4.14)

IT F(z) can be factorized as F = I(g), where I and 9 are entire functions with 9 being a nonlinear polynomial satisfying deg 9 < degP, then tPdz) and tP2(z)e P(z) have 9 as their common right factor.

155

Fix-points and Theory of Factorization

Remark. We note that the condition in Eq. (4.14) of the present theorem is weaker than that in Theorem 4.2, but here g is confined to be a polynomial of degree less than the degree of P. In order to prove the theorem we need to introduce some concepts, notations, and, lemmas about algebroid functions.

Definition 4.1. Let ao(z), adz), ... ,ak(z) be k+ 1 meromorphic functions (in the z-plane). A solution U = U(z) of the equation

is called a k-valued algebroidal function. Here we do not require that the polynomial P(U, z) be irreducible over the field of meromorphic functions. When all the aj(z)(j = 0,1,2, ... ,k) are polynomials, U(z) becomes the algebraic function. Let {e v } denote the countable set of the branch points of the algebroidal function U(z). Then, it is possible to draw from each point ev , a ray, Iv such that these rays will not pass any poles of U(z) and intersect one another. Let ru = uviv and Du = (lzi < oo)\ru. Then Du is a simply connected domain. In Du the function U(z) can be decomposed into k single-valued branches, i.e., Ut{z), U2 (z), ... Uk(z) - k analytic functions. If all the functions are distinct, and, moreover, given any two branches one could be obtained from the other by analytic continuation along a suitable path, then we shall call such an algebroidal function U(z) an exactly kvalued function. Otherwise, we call U(z) a degenerated k-valued function. H. Selberg introduced the standard Nevanlinna quantities, which are analog to the proximate function, counting function, and characteristic function in the Nevanlinna theory of meromorphic functions, for algebroidal functions as follows.

(4.15)

N( U) = ~ r,

jr n(t, U) - n(O, U) d + n(O, U)

kIt

t

k

I

og r

and

T(r, U) = m(r, U) + N(r, U) where k is the member of branches of the algebroidal function U(z).

(4.16)

156

Fix-points and Factorization of Meromorphic Functions

Clearly the "order" of an algebroidal function U(z) can be defined in terms of T(r, U) as in the Nevanlinna theory.

Lemma 4.3. (Selberg) Let U(z) be a 8-valued algebroidal function of finite order and let av(z)(v = 1,2, ... ,28 + 1) be arbitrary 28 + 1 distinct complex numbers. Then

2_+1 ( 1 ) T(r,U) = ?;N r'U(z)-a v +Ologr,

r-+oo

Lemma 4.4. (Prokopovich) Let J(z) be a transcendental meromorphic function of finite order with 6(e, I) = 1 for some value e (lei ::5 00). Let U(z) be an exactly k-valued algebroidal function satisfying T(r, U) = o(I)T(r, I) as r -+ 00. Then

N (r, J

~ U) ~

(1 - ~ + 0(1)) T(r, I),

r -+

00 •

We shall also need the following fact.

Lemma 4.5. (Prokopovich) Let g(z) be a transcendental meromorphic function satisfying N(r,g) = oT(r,g) as r -+ 00. If R(z) is a rational function of order k, and Q(z) is a non-constant meromorphic function satisfying T(r,Q) = o(I)T(r, g) as r -+ 00, the following inequality holds:

N (r,

R(J(z)~ _ Q(z)) ~ (k -

1 + o(I))T(r,

J),

r -+ 00 .

Now we proceed to prove Theorem 4.3. We will prove the case where 4>dz) is transcendental first. (No proof is needed if both 4>1 and 4>2 are polynomials!). Let us consider the equation

g(U) - z = 0 .

(4.17)

The equation defines some k-valued algebraic function. Furthermore it is easy to see that the k branches Udz), U2 (z), ... ,Uk(Z) are holomorphic in Du. Letting

4>(z) = J(g(z)) - 4>l(Z) , and substituting z by U(z) in the above equation, we have

4>(U(z))

=

J(g(U(z))) - 4>dU(z))

=

F(z) - 4>l(U(Z)) ,

(4.18)

Fix-points and Theory of Factorization

157

where 4>1(U(Z)) is an exactly s-valued analytic function (s ~ k). This means that the function 4>1(U(Z)) can be decomposed in the domain Du into s distinct single-valued branches in Du. For an arbitrary complex value a, which is not a Picard exceptional value of 4>dz), let {~j} denote the set of all the roots of the equation 4>1 (U) - a = 0 and Zj be the root of U(z) = ~j. It is evident, by Eq. (4.17)' that this equation has only one root. If we set IZjl = rj and g(z) = coz k + C1zk-1 + C2zk-2 + ... + Ck (co =I 0) . Then

rj = IZj) = Ig(U(zj)1 = Ig(~j)1

= (Icol + 0(1))k,-l k

as

j

--+ 00 .

This implies that we have the following approximation:

I~jl = W(Zj) I =

(Icol

+ 0(1))-t r},

J --+

(4.19)

00 •

We now examine the approximate values of the roots Zj of the equation ~ r}. It is easily verified that for r ~ ro (where ro is some sufficiently large number) any of the root Zj will satisfy the following estimation:

4>1(U(Z)) - a = 0 in the circle {izi

(4.20) where A is a suitable constant. Due to the one to one correspondence between the points ~j in {kl ~ Art} and the points Zj in {izi ~ r}, we have for r ~ ro

n(r'4>dU(~))-a) ~n(Art, 4>1(z~-a)

.

Hence, by virtue of (4.16)

N (r,

4>1(U(~)) _ a) ~ ~(1 + 0(1))N (Art, 4>dz~ _ a) ,r --+

00 •

Combining Lemma 4.3 and Eq. (4.14), we obtain

T(r, 4>1 0 U) ~

!; N

23+1

k

1

(

~ ( -; + 0(1)

)

4>1 0 U _ a" + O(log r)

r, )

!; N

23+1

= o(T(Art, fog)),

(

1) +

Art, 4>1(Z) _ a"

r --+

00 .

O(log r)

(4.21)

158

Fix-points and Factorization of Meromorphic Functions

Next we prove

T(Art, log) = O(T(r, j)),

r -+

00 •

Letting we have

w= z

(1 + ~ + ... + Ck k) t . CoZ CoZ

Since the radical tends to 1 as z -+ 00, it follows that in the region around 00, a single-valued branch of the radical function can be selected. Therefore w(z) becomes an analytic function in the domain {JzJ ~ ro}. Let Art> 2r. The image of the circle {JzJ = Ar t }, under the mapping w (that is 1 - 1 now), will be some curve "Ir lying in the ring:

for some positive constant d. Set

and

h(z) = I(g(z)) = I(coz k +

ClZ k - l

+ ... + Ck) .

Since a point Wo E "Ir corresponds to each point Zo E {JzJ = Art} so that h(zo) = 11 (wo), we deduce

T(Art,h):::; 10gM (Art,h) = log

Mhr, Id :::; log M(Art + d, Id ,

where Mhr,II) = maxJII(z)J. Using the well-known inequality between zE"lr

T(r, j) and log M(r, I), we have T(Art,/(g)) :::; 3T(2(Art + d), 11) . From T(r,/(cz k )) = T(JcJr k , j) and the fact that the Nevanlinna characteristic function T is an increasing function of r, it follows that

T(Art, I(g)) :::; 3T(Jc oJr k (Ar t + d)k, I) :::; 3T(Bl r,j), r -+ 00, where Bl is a suitable positive constant.

(4.22)

Fix-points and Theory of Factorization

159

On the other hand, we have

T(2Arf, J(g)) = T(2Arf, J2) ~

1

1

1.

3" log M(Ari", h) = 3" log M(-y,., il)

~ ~ log M(Arf 1

- d, Jd k

1.

= 3"T(lcol(Ark - d) ,j)

~ ~T(Arf ~

- d, Jd

1

3"T(B2r, J) , (4.23)

where B2 is a positive constant. By Eq. (4.14) and letting deg P

T(r, J(g))

= t,

we have

= T(r, F) = (1 + o(1))B3rt, r -+ 00

,

where B3 is a positive constant. Therefore,

It follows from Eqs. (4.22) and (4.24) that

T(r, j) = O(rt) as r -+

00 •

It follows from Eqs. (4.22) and (4.23) that

T(Ar f , J(g)) = O(rt) = O(T(r, j)), r -+

00 •

Combining this result and Eq. (4.21) yields

T(r,4>l(U)) =o(l)T(r,j) ,

r-+oo.

(4.25)

Let OJ be a zero of the function J(g(U(z))) -4>l(Z), and let IOjl = rj. Then = U(Oj) will be a zero of J(g(z)) - 4>t{z). It can easily be verified, from Eqs. (4.19) and (4.20), that Itjl ::; Art. Therefore

tj

n (Art, J(g) 1_ 4>J

~ n (r, J(g(U)) 1_ 4>l(U)) = n (r, J(z) _l4>d U ))

160

Fix-poin ts and Factorization of Meromorphic Functions

and hence

Thus, by an application of Lemma 4.5, we have (4.26) On the other hand, combining Eqs. (4.13)' (4.14)' and (4.24) yields

N (Art, f(g)l_

~ T(Arf, ~2)

~J

-

N (Art,

:J

+ 0(1) = O(l)T(Arf, f(g))

= o(l)T(r, J) . Comparing this result with Eq. (4.26)' we conclude that s ~l(U(Z)) is, in fact, an (single-valued) entire function. Set ~dU(z)) =


1. Hence

(4.27)

Then ~dU(z))

=
Choosing a point Zo such that g'(zo) ¥- 0 and g(zo) ¢ ru, we see that in a neighborhood of the point Wo = g(zo), there exists a single-valued inverse function g-l (w). Clearly, the function g-l (w) and one branch of the function U(w) are identical in the region of woo Let this particular branch be denoted as uio(w). Then we have

Since ~dU(z)) is entire, we have, for any i,j; 1 ~ i,j ~ k,

This means that not just in a region around Zo, but the whole z-plane

Fix-points and Theory of Factorization

161

From this result and Eq. (4.27)' we obtain (4.28) Let Then By Lemma 4.2, there exists polynomials Pl , P2 (deg P2 < degP = t) such that Consequently Note that the left-hand side of the above equation is an entire function of g, that is,

(4.29) We have therefore proved that

In view of this result and Eqs. (4.28) and (4.29)' we see that the theorem is proven for the case where
Case (i).
162

Fix-points and Factorization of Meromorphic Functions

Recall that Goldstein proved (in Theorem 4.2) that if Eq. (4.14) In Theorem 4.3 is strengthened by requiring that the orders of ¢Jl and ¢J2 be less than the degree of P, then the factorization F(z) = J(g(z)) for any two nonlinear functions J and 9 will lead to the conclusion that 9 must be a polynomial. Incidentally, while Goldstein and Prokopovich obtained their results, Gross and Yang derived a result similar to that of Goldstein and Prokopovich's. Their method was entirely different in that it utilizes Theorem 4.2, and some elementary properties of an algebraic function and its inverse function.

Theorem 4..4.. (Gross and Yang) Let P(z) be polynomial of degree m(~ 1), and let h(z)(¢ 0) and k(z)(¢ constant) be two entire functions of order less than m. Then h(z)eP(z) + k(z) is either prime or it can only be factorized as h(z)eP(z) + k(z) = J(L(z)) , (4.31) where L(z) is a nonlinear polynomial of degree nj J(z) = Il(z) exp[cz d ] + ,B(z) is an entire functionj a and ,B are entire functions of order less than mj and c is a constant ¢ O. Furthermore the following three relationships are satisfied: (i) nlm (namely ~ is an integer), (ii) h(z)eP(z) == a(L(z)) exp[cL(z)d], d = ~, (iii) k(z) = ,B(L(z)). Before going into the proof of the theorem, we need the following two lemmas.

Lemma 4..6. Let P(z,w) = Zn+alZn-l+a2Zn-2+ ... +an-w, where ai (i = 1,2, ... ,n) are constants, and write P(z, w) = TI7::1 (z - Zi(W)). Then every elementary symmetric function in Zi (w) is a polynomial in w. Proof. It is obvious. Lemma 4..7. Let w, zi(w)(i be an entire function. Then

=

1,2, ... ,n) be as in Lemma 4.6 and g(z) is an entire function in w.

E7:1 g(Zi(W))

Proof. Let us express

9 in its Taylor series: 00

g(z) = ao + alZ + a2 z2 + ... =

L

i=O

ai zi

163

Fix-points and Theory of Factorization

Then in any bounded domain D we have n

n

00

Lg(zd w )) = LLai[Z,;(w)ji ';=1

';=1 i=O

Since the infinite series is convergent absolutely in every bounded domain, we can rearrange the double summation and obtain

where the Pi

U = 1,2, ... ) are polynomials.

This proves the lemma.

Proof of Theorem 4.4. Suppose that F = he P + k is not prime. By Theorem 4.2 the only possible factorization for F is of the form:

he P where

f

+k

= f(L) ,

is entire and

L(z)

= ao + alz + ... + anz n ,

a,; constants,

(an "I 0) .

Let

P(z) =

Co

+ CIZ + ... + cmz m (cm"l 0)

(4.32)

and w

= L(z) .

We have, for sufficiently large Iwl (it will be assumed without loss of generality that all w's below are sufficiently large):

L(z) - w = an(z - zdw))(z - Z2(W)) ... (z - zn(w)) ,

(4.33)

where z,;(w)(i = 1,2, ... , n) are n-distinct branches (since, clearly, P(z, w) = L(z) - w is irreducible and of degree n in z). For i = 1,2, ... , n we have an expansion of the form (4.34) that is valid in

.1

00,

roots of unity, PI we have

=

where wi'

J.

= p,;wt, P,; (i =

1,2, ... , n) are n-distinct

J.

1, and wt is a fixed branch of w;. For Z,; (w) as above,

h(z,;(w)) exp(P(z,;(w)))

+ k(z,;(w)) =

f(w),

i

= 1,2, ... , n .

(4.35)

164

Fix-points and Factorization of Meromorphic Functions

Thus, for i

=1=

1, we obtain the following result

h(Zi(W)) exp(P(zi(W))) + k(z.(w)) _ 1 h(Zl(W)) exp(P(zt{w))) + k(zt{w)) - .

(4.36)

After dividing through the denominator and the numerator by h(zt{w)) x exp(p(z.(w))), the above quotient becomes

h(Zi (w))/ h(Zl (w)) exp(P(zi (w)) 1 + k(Zl (w))/h(zt{w)) exp( -P(zt{w))) P(Zt{w)) - k(z.(w)/h(zt{w))) exp(-P(z(w))) 1 + k(Zl(W))/h(zt{w)) exp( -P(zt{w)))

(4.37)

H we substitute Eq. (4.34) into Eq. (4.32), then ~

-~

P(z'(w)) = cm[wt (bo + L1wi

"

-~

+ L 2 wi " + ... )Im

+ Cm-l [wt~( bo + b-1Wi-~" + b- 2 Wi-~" + ... )Im- 1 + ... +Co -- dmWimin + dm-lWi(m-l)/n _.1.

+ ...

_--1-

+ do + d_ 1 Wi " + d- 2 wi

h

+ ...

(4.38)

We are going to verify that nlm by assuming the contrary nfm. Next, we choose a path of w, l, a straight line in the w-plane tending to infinity such that along l, dmw,;,/n = Idmw,;,/nl. Now from Eq. (4.38) we have

P(z'(w)) - P(zt{w)) = dm(pf _1)w,;,/n +dm-t{pf-l _1)wi m- 1 )/n + .... (4.39) H

nlm,

then

Re(pr - 1) < 0

(i

=1=

1) .

(4.40)

We now examine the behavior of the functions k(zt(w))/h(zdw))' and k(z.(w))/h(zdw)) along l. By assuming that h is an entire function of order less than m and by the property of the minimum modulus, we have for any given e > 0 and sufficiently large r,mh(r) =1= O(1)exp(-r m - e ), where mh(r) = minlzl=r Ih(re i8 )1. From this result, noting Eqs. (4.39), (4.40), and that k(z) is an entire function of order less than m, we see that

165

Fix-points and Theory of Factorization

I exp(P(zdw)) - exp(P(zdw))1 grows much slower than exp(-a:lwl m/ n ) as w -+ 00 for some constant a: > 0_ We conclude, after a simple verification, that the following three quantities:

[h(Zi(W))lh(zdw))] exp(-P(z, (w))) , [k(z, (w))lh(zdw))] exp(-P(zdw))) , and

[k(Zi (w))lh(Zl (w))] exp( -P(ZI (w))) all tend to 0 as w -+ 00 through a suitable sequence {w n } on 1. Therefore the left hand of Eq. (4.36) tends to 0 as w -+ 00 which is a contradiction. Hence, we must have nlm. Therefore, from Eq. (4.37), we have

P( Zj (W)) = dmW d

+ dm-lwi(m-l)/d + ... ,

where d = min is an integer. Substituting this into Eq. (4.19) for i 1,2, ... ,n and adding, we obtain

=

n

+ L k(Zi(W)) = nf(w) .

(4.41)

i=l Now, according to Lemma 4.7, E~=l k(zdw)) == T(w) is an entire function of order less than d (since it is easy to verify that each function k( Zi (w)) grows no faster than erd - j where e is small positive constant). It is also easy to verify that the growth of h(zdw)) (for i = 1,2, ... ,n) is no faster than that of erd - 6 ; j where Ci is a small positive constant. We also note that the function exp[dm_1w(m-l)/n + ... ] grows no faster than er-'lij where '1i is a positive constant (for i = 1,2, ... ,n), when Iwl is sufficiently large. Equation (4.41) can be rewritten as: C

;

n

L

h(zdw)) exp(dm_1w!m-l)/n + ... ) = [nf(w) - T(w)] exp( -dmw d )

•

i=l

(4.42) Clearly, the right-hand side of Eq. (4.42) is an entire function. Furthermore, the left-hand side by virtue of the above estimates grows no faster than

Fix-points and Factorization of Meromorphic Functions

166

erd.-~ for sufficiently large rj where '1 is a positive constant < d. Thus, we conclude that S(z) == [nf(w) - T(w)]exp(-dmw d) is an entire function of order less than d. Consequently,

f(w) == (S(w)/n) exp(dmw d) + T(w)/n . From this result and Eq. (4.31), we have

h(z) exp(P(z)) + k(z) == (S(L(z))/n) exp(dm(L(z))d) + T(L(z))/n . Hence

h(z) exp(P(z)) _ S(L(z)) exp(dm(L(z))d) == -k(z) + T(L(z)) n

(4.43)

n

Since T( w) and S (w) are entire functions in w of order less than d (= m/ n) and L(z) is a polynomial of degree n, it follows that both T(L(z)) and S(L(z)) are entire functions in z of oder less than m. Thus the right-hand side of Eq. (4.43) is an entire function of order less than m. Now the left-hand side can be expressed as

exp(P(z))[(h(z) - S(L(z))/n) exp(dm(L(z))d - P(z))] . To show that (h(z) - S(L(z))/n) exp(dm(L(z))d - P(z)) == U(z) == 0 we will suppose the contrary (U(z) ¢. 0) to be true. This means that k(z) - T(L(z))/n ¢. o. If dmL(z)d - P(z) == q(z) has a degree = m, then the function (S(L(z))/n) exp(dm(L(z)) - P(z)) will have three functions: 00,0, and h(z) as its deficient functions. This is impossible. Alternatively, q(z) can only be a polynomial of degree less than m. This implies that U(z) is of an order less than m. Therefore, the order of eP(zlU(z) is m (since k(z) -T(L(z))/n has an order less than m) which results in a contradiction. We must conclude U(z) == 0, and hence, from Eq. (4.43),

k(z) - T(L(z))/n == 0 or

k(z) == T(L(z))/n .

Fix-points and Theory of Factorization

167

Consequently we have

h(z) exp(P(z)) == (S(L(z))/n) exp(dmL(z)d) . Theorem 4.4 is proven.

Question. Why is the assumption k(z) ~ constant is necessary in the theorem? (Exhibit a counter example!) In the proof of Theorem 4.4 we used Theorem 4.2 to show that F(z) = + k is pseudo-prime and that a factorization of the form F = q(j) is impossible if q is a nonlinear polynomial and f is entire. Theorem 4.6 in the next section will also yield the result that he P + k is pseudo-prime. Application of the Tumura-Clunie theorem, allows for exclusion of the factorization of the form F = q(j). Rewriting F = he P + k = q(j) as he P = q(j) - k, leads to the impossible conclusion co(j - C1)n == he P ; where Co is the leading coefficient of q(z), n = degq, and c is a constant. In view of the above results, we draw the following conclusion.

he P

Corollary 4.2. Let Pdz)(i = 1,2, ... ,m) denote a polynomial of degree ti and let hdz)(i = 1,2, ... ,m) denote a non-constant entire function of order less than ti. Assume that 0 ~ t1 < t2 < ... < ... < t m ; m ~ 2. Then F(z) == ~m hi (z)eP;{z) is pseudo-prime. Moreover, if F is not prime, then the only possible form of the factorization of F is F = f(g), where f is entire and q is a nonlinear polynomial and a common right factor for all of the terms hie P ;, i = 1,2, ... ,m. 4.4. THE CRITERIA OF PSEUDO-PRIMENESS FOR ENTIRE FUNCTIONS To simplify the proof, we shall mainly deal with entire functions. In general, the steps to prove that a given transcendental entire function F is prime are (i) first prove that F is pseudo-prime (ii) prove that F cannot be expressed as F(z) = P(g(z)), where F is entire and P is a polynomial with deg P ~ 2, and (iii) prove that F cannot be expressed as F(z) = h(g(z))j where q(z) is a polynomial of degee ~ 2 and h an entire function. Now we introduce some sufficient conditions for determining the pseudoprimeness of a given entire function.

Theorem 4.5. (Goldstein) Let F(z} be a finite-ordered entire function with 6(a, F} = 1 for some complex number a. Then F is pseudo-prime.

168

Fix-points and Factorization of Meromorphic Functions

(The function F(z) necessary one)_

= ee·

shows that the restriction on the order of F is a

Proof. We may assume, without loss of generality, that a assumption we have

K(F) = lim {N(r,O, F) r-+oo

= o.

By the

+ N(r, 00, F)}/T(r, F)

- . N(r,O,F) -1. N(r,oo,F) + 1m T(r, F) r-+oo T(r, F) 1 - 15(0, F) + 1 - 15(00, F) = 0 .

< hm

-

=

r-+oo

(4A4)

Using a result of Edrei-Fuchs' (Gomm. Math. Helv. 33 (1959), fA, 258295), it follows that the lower order of F satisfies p - ~ ~ p. < p + ~ for some positive integer p. In this case the following result holds: To any finite ordered entire function F satisfying conditions of Eq. (4A4) there exists a sequence of circular arcs bj}~1> i.e., 1j lying on the circle {Izl = rj} with rj i 00. Moreover, the angular measurement of 1j is ~ 21r/3p for i = 1,2, ... so that when z E A(= U1j) and Izl ~ ro (ro is sufficiently large) the following estimation holds:

(4A5) In the same paper, Edrei-Fuchs also proved that these exists a sequence of segments {lj}~l on which F also satisfies the inequality in Eq_ (4-45), where lj defines a segment with one end point as rje i8j E 1j and the other end point as rj+1ei8j+l E 1j+1, i = 1,2, - - .. Therefore bj}~l and {lj}~l form a continuous curve L which Eq. (4-45) holds when Izl is sufficiently large. If F were not pseudo-prime, then we would have F(z) = J(g(z)) where both J and 9 are trancendental entire functions. According to a theorem of Polya's (noting that F has finite order)' J is of zero order. Hence, there exists a real sequence {Rn} with Rn i 00 such that min If(z)lIzl=R"

00

as

n-

00 .

(4-46)

But by Eq. (4-45) F(z) - 0 as z tends to 00 along L. Therefore we must conclude that r = g( L), that is, the image of Lunder 9 is a bounded curve, otherwise, Eqs. (4-45) and (4-46) will yield a contradiction_ Next we show

Fix-points and Theory of Factorization

00 along L, where 0: is a zero of f(z). Let be the equation of the curve L, h(t) -+ 00 as t -+ 00. can be represented as

that g(z)

-+

0: when

z = h(t)(O $ t < Then

r

169

z tends to

00)

U = g(z) = g(h(t)) ,

0$

t<

00 •

Since r is bounded, there exists some closed disc K = {z liz I $ R} containing r. Without loss of generality, we may assume that f does not vanish on Izl = R. Let O:i' j = 1,2, ... ,k, be all the distinct zeros of f in Izl < Rand 5 = mini,i=1,2, ... ,k{di, 100i - O:il} where di defines the distance i¢k from O:i to the complement set of K. Let m be any positive number and Gi denote the circle Iz - O:il = 5/m. Letting minzEc; If(z)1 = 1110(> 0) and mo = min{mI,m2, ... ,mk,minzER If(z)I}, we have mo > o. Therefore, If(z)1 < mo for some z E K, implies that the point z must li( outside the circles Gi , i = 1,2, ... ,k. When t ~ to then g(h(t)) E K and If(g(h(e))) I < m. This means that when t ~ to, g(h(t)) lies inside some circle Gi . The set {g(h(t)) It ~ to} is connected. It follows from this and the fact that the sets Gi(i = 1,2, ... ,k) are mutually separated that there exists some positive integer, j(1 $ j $ k) such that the following inequality holds:

Ig(h(t)) -

O:il <

5

(t

-, m

~

to) .

Since m can be arbitrarily large, this implies that g(z) -+ 0: as z --+ 00 along L, where 0: is a zero of f. Therefore we have, given anye > 0 and z E L with Izl ~ ro,

Ig( z) - 0: I $

Vz E ,i, j ~ jo .

e,

(4.47)

Assume that the multiplicity of the zero point 0: is s(~ 1). It follows that there exists a positive constant A (> 0) such that whenever Iz - 0:1 $ e,

If(z)1 ~ Alz that is, whenever Ig(z) -

0:1

0:1" ,

(4.48)

$ e,

If(g(z))1

~

Thus, for z E 'i,j ~ jo (or z E L,

IF(z)1

~

Izl

Alg(z) -

0:1" .

(4.49)

~ ro)

Alg(z) -

0:1" .

(4.50)

170

Fix-points and Factorization of Meromorphic Functions

Since the inequality in Eq. (4.45) holds for z E "ti' we obtain 8

log Ig(z) -

0:1

+ log A

11"

~ - 16 T(lzl, F) .

Consequently, for z E 'Yi(j ~ 30), we have log +

I z 1- I g

()

~ log

0:

I

g

()1 Z

-

0:

I

11" -T(lzl, F) + -log8 A . 168

~

When i ~ 3"a with z = re iO from the above results and by applying Nevanlinna first fundamental theorem, we obtain, for an integer p > 0

T(ri' g)

+ 0(1)

~ m(ri' 0:, g)

> -1

!

- 211"""i ~

Since T(r, F)

-+ 00

as r

1 1211" log+ I (-0) 1 = -211" 0 g reI

log+

-+ 00,

0:

IdB

Ig(re'O)_1 - IdB 0:

11" -1 ( -T(r-,F) 3p

-

168

J

A) .

log+8

(4.51)

the above equation yields

T(ri' g) ~ BT(ri' F),

i

~

30 ,

( 4.52)

where B is a suitable positive constant. However, according to a theorem of Clunie's, for any two transcendental entire functions g and f, lim T(r, f(g)) = r-+oo

T(r, g)

00 .

(4.53)

This contradicts with Eq. (4.52). We must conclude that it is impossible for both f and g to be transcendental entire in the factorization F = f(g). This completes the proof. Remark. Goldstein remarked that Theorem 4.5 remains valid under either of the following two conditions: (i) 6(0, F') = 1 (ii) Ea;o!oo6(a,F) = 1. It was also remarked that the Edrei-Fuchs' result applies not only for 6(a, F) = 1 but also for 6(a, F) > 1- e(p), where e(p) is a positive constant

171

Fix -points and Theory of Factorization

(0 < e(p) < 1) depending on the order of F_ The above remarks also lead to an interesting conjecture as follows_

Conjecture 2. (Fuchs) Let F be an entire function of finite order. IT 8(a, F) > 0 for some complex number, then F is pseudo-prime. Using an argument similar to that used for the preceding theorem the following result can be obtained.

Theorem 4.6. (Gross and Yang) Let P(z) be a polynomial of degree t (~ 1) and hl(Z) and h2(Z)(t 0) be two entire functions of order less than t. Then F(z) == h 1 (z)e P(z) + h2(Z) is pseudo-prime. Hint: Write F(z) as h2(z){~!!=leP(z) + 1}. Question. Does the theorem remain valid if only T(r, hd r -+ oo,i = 1,2 is assumed?

= o(1)T(r, eP )

Recall that a transcendental entire function F is called left-prime or E-Ieft-prime if F = f(g) with f and g being entire implies that f must be linear whenever g is transcendental. F is called right-prime or E-rightprime if F = f(g) with f and g being entire implies g must be linear whenever f is transcendental. Clearly we have (i) IT E is both right and left-prime, then F is E-prime. (ii) A left or right-prime transcendental entire function must be a pseudo-prime. We now provide some criteria for left-primeness.

Theorem 4.1. (Ozawa) Let F(z) be an entire function of finite order whose derivative F' (z) has an infinite number of zero. Suppose for any complex number c, the following simultaneous equations:

{ F(Z)=C F'(z) = 0

(4.54)

have only a finite number of solutions. Then F(z) is left-prime.

Proof. Suppose that F has the factorization F = f(g)j with f and g being transcendental entire functions. From Polya's theorem we must have p(F) = p(F') = O. Hence f'(d has an infinite number of zero, that can be summarized as {~i }~1' There must be some fixed ~i such that the solution to the equation g(z) = ~i are an infinite set. Let {Zn}:=l be the set. The

172

Fix-points and Factorization of Merom orphic Functions

simultaneous equations

{ F(zn) = J(g(zn)) = J(~j) = c F'(zn) = J'(g(zn))g'(zn) = 0,

n

=

1,2, ...

have an infinite number of solutions. This is a contradiction to the hypothesis. We conclude that F must be pseudo-prime. Assume that F = P(g)' where P is a nonlinear polynomial and g is an entire function. P'(d has at least one zero, ~. IT g(z) = Q results in an infinite number of solutions, then using the same argument as above we will get a contradiction. Now assume that g(z) = Q only has a finite number of solutions, this results in

g(Z) =

Q

+ Q(z)eq(z)

,

where q and Q are polynomials. This gives g'(z) a finite number of zeros. Since the assumption states that F'(z) = P'(g(z))g'(z) has an infinite number of zeros, it follows that there must exist a root of P'(~), {3, not equal to Q, such that g(z) = {3 has an infinite number of solutions. This will again lead to a contradiction. The theorem is thus proved.

Exercise. Prove that F(z) = eZ + P(z), where P is a polynomial, is left-prime. Use this result to show that F is E-prime. Exercise. Illustrate the requirement that F'(z) has an infinite number of zeros is a necessary condition for the validity of Theorem 4.7. When no restriction is imposed on the order of F(z), the following results.

Theorem 4.8. (Ozawa) Let F(z) be a transcendental entire function with N (r, },) ~ kT(r, F) for some positive constant k. IT for any complex number c, the system of Eqs. (4.54) has only a finite number of solutions, then F is left-prime. Proof. Suppose that F = J(g), where J and g are both transcendental entire. Finally we assume that J' (~) = 0 has no roots at all. Then

N (r, ;,) = N (r, :,) ::; T(r, g') + 0(1) ::; (1 + e)T(r, g), n.e.,

(4.55)

where "n.e." means the inequality holds nearly everywhere for sufficiently large values of r except possibly a set of r values of finite length.

173

Fix ·points and Theory of Factorization

On the other hand, for any positive integer p and some constant A (not a Picard exceptional value of J), we have

T(r, F)

~ N (r, F ~ A) + 0(1) ~

t

N (r,

3=1

~

(p - 1)T(r, g)

g! 0') + 0(1) , 3

+ O(log rT(r, g))

n.e. ,

(4.56)

where OJ E 1-1 (A). The combination of Eqs. (4.55) and (4.56) yields a result that will contradict the hypothesis of the theorem: N (r, ;,) ~ kT(r, F). If we assume that I has only one zero, ~o, and g( z) = ~o has a finite number of roots, it follows that

N (r, ;,)

= N (r, ;,) = O(log r) :::; T(r, g') + O(log r) :::; T(r, g) + O(log rT(r, g))

n.e ..

This leads to the same contradiction found in the previous case. Alternatively we assume that I'(d has only one zero, ~o, but g(z) = ~o has an infinite number of roots, {Zj}. Then the following simultaneous equations

{ F(z) = I(~o) F'(z) = 0 have an infinite number of solutions {Zj}. This is also a contradiction to the hypothesis. Now we assume that f'(d has at least two distinct zeros. By choosing one of the roots, ~1 so that g(z) = ~1 has an infinite number of roots, we will arrive at the same contradiction. This also proves that F is a E-pseudoprime. Finally, we assume that F = P(g), where P is a polynomial and g is a transcendental entire function. If P' (~) has only one zero and g(~) = ~o has a finite number of roots, then

g(z)

= ~o

+ Q(z)eG(z}, g'(z)

=

(Q'

+ G'Q)eG(z)

where Q is a polynomial and G is an entire function. Then

N (r,

-i) : :; N (r,

Q' +1 G'Q) =

oT(r, g)

n.e ..

(4.57)

174

Fix-points and Factorization of Meromorphic Functions

IT on the other hand, t = degP, then there exist some arbitrarily small positive number e and e' such that

N (r, ; ) = N (r, ;,) + O(log r) ~

(1 + e)kT(r, F)

~

k(t - 1)(1 + e') T(r, g) .

This will contradict Eq. (4.57) unless t = 1, i.e., P is a linear polynomial. The cases, like P'(s") = 0 can be proposed as having a root, s"o, such that g(z) = S"O has an infinite number of roots, or P'(s") can have at least two distinct zeros and one of them, s"b can enable g(z) = s"l to have an infinite number of roots, etc.; can be argued as before and similar contradictions will result. This also completes the proof of the theorem.

Remark. Noda noted that the condition N (r, j.,) > kT(r, F) of the theorem can be replace by either (i) requiring N (r, j.,) ~ kT(r, F) on a set of r values of infinite measure for some k > 0 or (ii)

N (r, ;, ) - [N (r,

~)

- N (r,

~ ) ] ~ kT (r, ~),

n.e ..

These two facts are useful in the proof of Theorem 4.9 found in the next section. 4.5. THE DISTRIBUTION OF THE PRIME FUNCTIONS We would like to know like the distribution of prime number r in the set of integers; the distribution of prime functions in the family of entire functions. In the section we shall resolve two related questions: (A) (Gross) Given any entire function I, does there exist a polynomial Q such that 1+ Q is prime? B) (Gross, Osgood and Yang) Given any entire function I, does there exist an entire function g such that gl (the product) is prime? Noda provided affirmative answers to the above two questions as follows.

Theorem 4.9. (Noda) Let I(z) be a transcendental entire function. Then the set {ala E CD and I(z) + az is not prime} is at most a countable set in the complex plane
Fix-points and Theory of Factorization

175

is a countable set E of complex numbers such that the simultaneous equations { f(z) - az = c

(4.58)

f'(z)-a=O

have no more than one common root for any constant c(E CD) provided that a belongs to CD/E.

Proof. We write A

= CD\{z E CDI!"(z) = O}

.

Clearly A is an open set. It is easily verified that one can choose an open covering of A, say {Gi } ~ l' such that the following three conditions will be satisfied: 00

UG

(i)

i

=

A,

i=l

(ii) f'(z) is univalent in Gi(i = 1,2, ... ) and (iii) Di = {J'(z)lz E Gi } is a disk (i = 1,2, ... ). Set

F(z) = f(z) - zf'(z) .

(4.59)

and define functions Ui and v.:(i = 1,2, ... ) as follows:

Udw) namely

= (f'IGd-1(w) (w

f 0 Ui(w) =

W

E Di ,

i

=

(4.60)

1,2, ... );

Vw E Di,

(4.61) Let

1= {(i, i) I with i, i positive integers such that Di n Di and Vi(w) ¢ Vi(w) (w E Di n Din S.i

= {wlw E

=1=