Baxma, I.N.
Math. Annalen, Bd. 129, S. 174---180 (1955).
The Iteration of Entire Transcendental Functions
and
the Solution of the Functional Equation f{/(~)} = F(~).
By
I. N. BAKER in Adelaide.
1. Introduction.
Let F (z) be an entire analytic function. Define the n-th iterate of F(z):
(1)
F0(z ) = Z, / ~ + I ( Z ) : /~ {--~(Z)} : / ~ n { F ( z ) } ,
• = 0, l, 2 . . . . .
These iterates of positive integrM order are entire functions satisfying the
functional equations
(2)
F,,,{Fn(z)} = F,~{F,,(z)} --- Fm+n(z)
where m and n are positive integers. One may also introduce the iterates of
negative integral order by writing F_~ (z) for the inverse function of F ~ (z).
Thereby many-valued functions are introduced but the equations (2) remain
valid for a suitable choice of branches.
SCn~6DER [1], [2] originally studied the "natural" iterates (i. e. those of
integral order) with a view to finding for F n (z) a closed analytic expression
containing n as a parameter. The iterates of non-integral order then appeared
simply by assigning non-integra/values to this parameter. A modern work
which treats iterates of "complex order" is that of T6PFER [3]. JULIA [4] and
FATOU [5], [6] have made a thorough investigation of the natural iteration of
rational and entire functions, relying on MONTEL'S theory of normal sequences
of analytic functions.
In the following we shall be interested in (analytic) solutions/(z) of the
functional equation
(3)
/ {/(z)} = F(z), F ( z ) entire.
The problem which forms the basis of this paper is:
W h a t restrictions must be placed on F(z) if there are to be entire solutions /(z) ?
The main theorem 1 bears directly upon this point, and in section 3 we
discuss some eases not covered by the theorem. In the final section some applications of theorem 1 are made, in particular to discussing further K ~ s E R ' s [7]
treatment of the case F (z) --- ez.
2. The Case Where F(z) is o! Finite Order and Bounded
on a Curve Leading to Infiniiy.
Theorem 1. I f $'(z) is an entire function of finite order bounded on some
continuous c u r v e / ~ which extends to infinity, then the functional equation
(3)
haa no entire solution.
/
{/(z)} =
F(z)
On Entire Transcendental Functions.
175
Proof: I f f(z) is an entire solution of (3) it must be of zero order, as follows
from the following result of FATou [6, p. 343]:
Let E(z), F(z) be two entire transcendental functions. On [zI ----r the
maximum modulus of E(z) is M(r), that of E {F(z)) is Ml(r ). Then however
large we choose the positive constant q, we have for certain arbitrarily large
values of r:
Ml (r) > M (rq) .
I t follows that if a > 0 be the order of f (z), the order of F (z) is greater than
q ( a - e) however large q and however small e; i. e. F (z) is of infinite order,
against assumption.
Now consider the behaviour of f (z) o n / ' . Either f (z) is bounded o n / ' or
it maps /" on to a continuous curve Z which extends to infinity. But F(z)
= f {f (z)} is bounded o n / ' so that in the latter case f (z) is bounded on Z, and
in either event f {z) is bounded on a continuous curve extending to infinity,
which is impossible for an entire function whose order is less than ,~ - - one
recalls the result [8, p. 274] that for a function g(z) with this property there
is a sequence of values of r tending to infinity through which
min Ig(z)l-~ oo.
I~1 =r
Corollary. I f F(z) is of finite order and there is an exceptional value A which
F(z) takes at most a finite number of times, then (3) has no entire solution.
For such an exceptional value is an asymptotic value [9, p. 34] and on the
curve extending to infinity along which F (z) tends to ,4, F (z) will be bounded.
3. Cases not Treated in the Last Section.
I t is obvious that there are functions F(z) such t h a t f (z) m a y be entire:
exp (exp z) is such a function, but its order is infinite. The question arises
as to whether this is possible when F(z) is of finite order, and this is answered
by the provision of examples which demonstrate the possibility. I n fact we
can prove the
Theorem 2. There exist entire functions ]l(z), fz(z) and f3(z) of zero order
such that
(i)
(ii)
(iii)
fl{fl(z)} is of zero order
fa{fp.(z)} is of infinite order
f3{f3(z)} is of finite non-zero order.
Remarks: Where it is necessary to check the order of the functions introduced below one uses the result that if ~ is the order of the function ~ a~z n,
it=0
one has [10, p. 40]
1 _ lira -- log la~l
Q n~---~ nlog n
Use is made of the notations with respect to f(z) = X anzn:
176
I . N . BAKZa:
/x(r):
(r) :
M(r):
M'(r):
the
the
the
the
maximum term of the sequence ]anl r n.
central index, the index of the maximum term.
maximum for ]zI --- r of [](z)l.
maximum for Izl =- r of If {J(z)}l.
It is to be noted that these four functions are monotone increasing.
We reserve square brackets to have the meaning: [A] is the greatest integer not exceeding A.
The discussion depends on the lemma:
(4)
#(r) < M(r) < ta(r) {1 + 2 v r + ~
,
proved in [9, p. 32] and the fact that we deal with functions for which the
coefficients an are real and positive, so that
M(r) = f(r), M'(r) = f {/(r)} = i {M(r)}.
Proof of Theorem 2.
(i)
Take fl (z) =
I 1 +loge ]
r i l l
0
exp (n')
r
1 +logr
2
A simple consideration of (4) shows that for all large enough r
M(r) < exp {1_ (logr)~}
and
M' (r) < exp {~- (logr)a),
so that the order of fl{fl(r)} is 0.
(ii)
Take f~ (z) =
1
exp {n (log n)'}"
Now for continuous n, e x p { n l o g r - - n ( l o g n ) ~} has its maximum where
n = n(r) = exp (--1 + V 1 + log r) > exp (~- V l o g r ) for r > e~.
Certainly
/~(r) ~ exp {[n(r)] log r - [n(r)] (log [n(r)]) 2}
exp {(n(r) - - 1) (log r - - log n(r). log n(r))}
> exp {exp (-~-V~10gr)} for sufficiently large r.
From (4)
M'(r) > g {g(r)} > exp {exp J- fexp
which is not bounded by any expression of the form exp(r x) and shows that
]s{fa(z)} is of infinite order.
(iii) We deal now with the comparatively intricate problem of constructing
such an f.,(z) that fs{f3(z)} has finite non-zero order. The first step here is to
On Entire Transcendental Functions.
177
produce a suitable real (nonanalytic) function ~ ( r ) satisfying ~ {qb(r)} = e'.
B y taking a series with positive coefficients, for which p ( r ) is fairly close
to q5 (r), a satisfactory ]a (z) is obtained as is suggested b y (4). Firstly t h e n we
p r o v e the
Lemma: There is a continuously differentiable function qS(r)such t h a t
~(qb(r)) = e ~ and the four functions ~b(r), ~ ' ( r ) , ~ (rr} and ~r~'(r) all t e n d
m o n o t o n e l y (for r > 1) to infinity with r. W e introduce the notation: I 0 is the
interval 0 ~ r ~ 1. I n general I n is defined inductively: ln+1 is the image of
I n under the m a p p i n g x -+ e ~. ~ In covers the positive real line. W e assert
n=0
t h a t the conditions of the l e m m a are satisfied b y the function defined in I 0 b y
¢ ( r ) = r + 2z
O ~--r ~ l }
= e x p ( r - - {-)
1 _< r <
a n d defined in I n generally b y repeated application of the relation
(5)
qb (e ~) = exp {q5 (r)}.
Thus, for example, in I1:
qb(r) = e 11, r,
1 =< r ~ e'l' /
e x p ( e -'l' r), e 'h • r g e I •
Obviously in I 0 : q 5 (qb(r)} = e r, and it t h e n follows from (5) t h a t this is so
in e v e r y I n.
N o w the four functions mentioned in the l e m m a are all continuous a n d
m o n o t o n e increasing in 11 . Using the m e t h o d of induction, suppose this is
true in a n y I~. Differentiating (5) gives
er ~'(e~)
q~(~)
r q~'(r)
so t h a t - ~ - ~ - i s
-
~)' ( r ) ,
continuous a n d m o n o t o n e increasing in I~+ v
A similar
result follows for the other functions from the identities
• (e~)
e~ - exp {~5 (r) - - r}
and qb'(e r) = q Y ( r ) e x p {qb (r) - - r}. Since the functions increase m o n o t o n e l y
with r, t h e y either t e n d to infinity or are bounded a n d t e n d to a finite limit.
I n each case the latter supposition conflicts with the fact t h a t q5 {q5 (r)} = e ~.
r~'(r)
Thus for e x a m p l e i f - ~ ( r ~
tends to 1 as r tends to infinity:
r = ~'~-~~ (~") - ~ ¢"{'~(")}
~{~(~)} ~ ( r ) -a~'(,.)
~ - ( ~ - - * l~
as r ~ ~ , since q5 (r) -~ co m o n o t o n e l y with r; b u t this is obviously false.
I f qS(r), ~'(r) or ~--~rr) are bounded t h e n certainly there is a K such t h a t
qS(r) < K r a n d e r = ~b{qS(r)} < K ~ r which is impossible. This completes t h e
proof of the lemma.
178
I.N. BAKER:
The Junction Is(z). The functions/~(r), ~(r) defined above are connected
by the relation
(6)
log/~(r) = log laol + --j Y(:)-- dx
0
where a o ~ 0 is the constant term in the expansion of the relevant entire
function. ~(r) is a (monotonely increasing) step function with a denumerably
infinite set of isolated discontinuities rn > 0, n = 1, 2 . . . . . at each of which ~ (r)
increases b y an integer. As VALIRON [10, p. 28--30] shows, to any such ~(r)
given, in advance one can construct a number of entire functions which have
~(r} as their central index function. For a given choice of ta01 ~= 0, it is possible
to find one of these whose coefficients are real and positive and which is
in fact a majorant of all similar functions. Having chosen ~(r) and a o it is
this function which, for definiteness, we shall use for/a (z).
We choose for ~ (r):
[ rO'(r) ]
(r) = 1 + [ - ¢ ~ ] ,
=0,
=2,
r _~ 1
0<r<0.1
0.1<r<l
Then
/v(~)
dx > /
--~-~dx
¢'(x)
0
and
f
r
dx > j
0
O(x) dx, r > e .
0
When one makes a 0 = {, (6) becomes
log ~,(r) --- log ~ +
0
f ~
a~ > log ~ +
/ ~'(x)
- ~ i ~ a~ > log ~(r),
r~>e.
0
Then M'(r) >/~ {/~(r)} > • {O(r)} = e r. But on the other hand
/-¢~+~/ax,
1
so that for some positive K
/~ (r) < K r • (r).
For all r > 0 . 1 , ~(r) > 1, so t h a t from (4) and the line above:
1) < 3Kr2
M(r) < K rtl)(r) {1 Jr 2 + 2(r q- 1) O'(r-lO(r+ l) }
for large enough r.
N o w • {O(r)} = e" = ~
(e') =
O'{O(r)}. O'(r)
~'(r-t-1) ~ ( r )
O(r+l)
On Entire Transcendental Functions.
179
from which one sees t h a t ~ ' (r) < ~5(r) for all r > 1 ; so that:
(7)
M (r) < 3 K r2qS(r).
However small ~ > 0 m a y be,
+o)#(r) } = (1 + ~)r a+n #(r-V~)~l~'
+~)
(1
r ~d- .{l°g #(rl
r#(r)#(r)
tends to infinity with r, and so remains greater than arbitrarily large N for
r > r(N). A simple integration shows that there exists a positive real C
such that
¢ ( r 1+~) > Or'Vff)(r), r > r(N).
With this result (7) becomes
(8)
M ( r ) < ~ ( r 1+~)
~'(r)
Since r ~ ~ oo with r it follows that for large r,
r < {~ (r)}~
however small ¢¢ > 0 is chosen, and replacing r by q~(r):
From this result, (7) and (8) it follows that
M'(r) = M {M(r)} < 3 K {qS(rl+e)} ~ q5 {qS(rl+e)}
< 3 K exp {rl+e + 2~¢ rl+e}.
But ~¢ and (~ may be chosen arbitrarily small, so that the order of/3{~8(z)} is
exactly 1. The order of ]3 (z) can only be zero.
4. Consequences of Theorem 1.
We now apply the main theorem to derive a property of analytic solutions
of (3).
In the iteration of an entire function F (z) there arises a set ~ (F) which is
of great importance: ~ (F) is the set of points z where the iterates F~ (z) of
positive integral order do not form a normal family in the sense of MDNTEL.
The term " F a t o u exceptional point" is used to denote a point a such
that the equation F(z) = ¢¢ has no solution in z, except possibly z = a. There
can be at most one such point.
FATOTY[6] has investigated the set ~ (F) thoroughly and has proved the
following result:
Let c be a small circle surrounding z 0 ~ ( F ) , and cn its image under the
transformation z -~ F n (z). I f A is any bounded region of the plane excluding
a small circle d about the Fatou exceptional point if this exists, then for
sufficiently large n, cn will cover A.
From this follows
Theorem 3. I f f(z) is an analytic solution of (3), where F(z) is bounded on
some curve extending to infinity a n d / ( z ) is analytic at any one point o f ~ ( F )
and also at the Fatou exceptional point if this exists, then ](z) cannot be
single.valued.
180
I . N . B~K~R: On Entire Transcendental Functions.
Proo/: S u p p o s e /(z) s i n g l e - v a l u e d a n d a n a l y t i c a t zoEqj(F ). T a k e for c
a circle s u r r o u n d i n g z 0 in which t h e e x p a n s i o n of ](z) a b o u t z 0 converges.
T h e n / (z) m a y be c o n t i n u e d i n t o a n y c, b y r e p e a t e d a p p l i c a t i o n o f t h e r e l a t i o n
/ { r ( ~ ) } = / [ / { / ( z ) } ] = F {/(z)}.
I f a F a t o u e x c e p t i o n a l p o i n t ~ exists, t a k e a circle d a b o u t ~ in which /(z) is
a n a l y t i c . T h e n ff D is a n y b o u n d e d region, o f t h e p l a n e o u r p r o c e d u r e o f
c o n t i n u a t i o n shows t h a t ](z) is a n a l y t i c in D - - u , a n d hence i n D ; t h a t is
](z) is entire, a g a i n s t t h e o r e m 1. q.e.d.
T h e i n s t a n c e o f F (z) = e~ is o f p a r t i c u l a r i n t e r e s t a n d h a s b e e n t r e a t e d b y
K~ESER [7], who has c o n s t r u c t e d a solution ](z) of (3) which on t h e whole
real axis is a n a l y t i c a n d real. e~ has t h e e x c e p t i o n a l v a l u e 0, a n d T6PFER [11]
has s h o w n t h a t e v e r y p o i n t of t h e real axis belongs to ~ (eZ). I t follows t h a t
KN~.SEI~'S solution is n o t single-valued.
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(Eingegangen am 6. September 1954.)