1. No category

Computational Methods and Function Theory
Volume 9 (2009), No. 1, 57–64
How to Detect Hayman Directions
Andreas Sauer
(Communicated by Ilpo Laine)
Abstract. We introduce the class of Hayman exceptional functions which
seem to play a similar role for the study of Hayman directions as Julia exceptional functions do for Julia directions. Further we prove that for every
b there exists n ∈ N such that
transcendental meromorphic function f : C → C
f (k) has a Julia direction for every k ≥ n.
Keywords. Hayman direction, Julia direction, filling disks, cercles de remplissages, singular direction.
2000 MSC. 30D35.
1. Julia directions
b
In [5] Julia proved that for transcendental meromorphic functions f : C → C
with an asymptotic value, Picard’s Theorem holds on relatively small subsets of
the plane. Julia considered a quite general class of sectorial neighbourhoods of
paths to ∞ in C, but most of the following research on this topic was restricted
to the case of sectors. It is standard to call a ray J from 0 to ∞ a Julia direction
if, in every sector ∆ containing J, Picard’s Theorem holds for f , that is, f takes
b with at most two exceptions infinitely often in ∆.
every value of C
With the help of the theory of normal families (see [9]), in particular Montel’s
Theorem and Marty’s Theorem, Julia’s proof yields the following result, involving
the spherical derivative f # := |f 0 |/(1 + |f |2 ).
b with a sequence zj → ∞
Theorem 1. Every meromorphic function f : C → C
such that
(1)
|zj |f # (zj ) → ∞
possesses a Julia direction.
Received October 31, 2007, in revised form January 21, 2008.
Published online February 26, 2008.
c 2009 Heldermann Verlag
ISSN 1617-9447/$ 2.50 58
A. Sauer
CMFT
The existence of such a sequence zj is the most common criterion for the existence
of a Julia direction. The points zj serve as centres of so called filling disks (or
cercles de remplissage) for f ; see [15] for the exact definition.
Hence meromorphic functions without Julia directions necessarily satisfy
1
#
(2)
f (z) = O
|z|
for z → ∞. Transcendental meromorphic functions satisfying (2) exist and are
called Julia exceptional functions. A. Ostrowski considered this class of functions in his classical paper [6]. It contains a complete characterization of Julia exceptional functions in terms of the distribution of zeros and poles of f
[6, Satz 5]. The following example of a Julia exceptional function is also given
there [6, p. 258]:
∞ n
Y
q −z
(3)
F (z) =
qn + z
n=0
where q > 1. This function has indeed no Julia direction. An elementary proof
of this fact can be found in [10].
Condition (2) implies that f grows very slowly: for the unintegrated AhlforsShimizu characteristic
Z
1
(f # (z))2 dz
A(r, f ) =
π |z|≤r
it follows that A(r, f ) = O(log r), and thus for the Nevanlinna characteristic that
(4)
T (r, f ) = O((log r)2 ).
In this sense “most” transcendental meromorphic functions posses a Julia direction. Furthermore, this growth condition is sharp since the growth of T (r, F ) for
the function F in (3) is of the same magnitude as (log r)2 .
2. Hayman directions
The starting point of our investigations is Hayman’s classical theorem [4, Cor.
to Thm. 3.5] on the value distribution of f and its derivatives:
b takes every
Theorem 2. Every transcendental meromorphic function f : C → C
value of C infinitely often or every derivative f (k) takes every value in C \ {0}
infinitely often.
Now it is natural to consider the question, whether there exist directions such that
Hayman’s Theorem is true in every sector containing the direction. Directions
with this property are called Hayman directions and have been considered by
several authors (see [14, 15, 2, 3, 13]). The basic theorem was proved by Yang
Lo [14].
9 (2009), No. 1
How to Detect Hayman Directions
59
b be a transcendental meromorphic function with
Theorem 3. Let f : C → C
(5)
lim sup
r→∞
T (r, f )
= ∞.
(log r)3
Then f possesses a Julia direction that is a Hayman direction.
The fact that the Hayman direction detected in Lo’s Theorem is also a Julia
direction was not explicitly stated in [14], but it follows directly from the proof.
Further an example of Rossi [7] shows that the growth condition (5) for this
formulation of Lo’s Theorem is sharp:
(6)
G(z) =
∞ Y
q
n=0
√
n
√
q n
−z
+z
with q > 1. For this function, T (r, G) ∼ (log r)3 /3, it has exactly two Julia
directions (the positive and negative imaginary axis), but neither of them is a
Hayman direction.
This does not mean that G has no Hayman direction at all. The only candidate
for a Hayman direction is the negative real axis, though. In small sectors around
the positive real axis G is bounded. Around other directions except the negative
real axis G omits 0 and ∞ and grows like a polynomial, which together with
Cauchy’s formula shows that from some k on all derivatives G(k) are bounded in
slightly smaller sectors. (All these properties of Rossi’s example follow from the
analysis of that example given in [7] and [2].)
There is also a condition for sequences similar to the one in Theorem 1 (see [3]):
b with a sequence zj → ∞
Theorem 4. Every meromorphic function f : C → C
such that
(7)
|zj |f # (zj )
→∞
log |zj |
possesses a Julia direction that is a Hayman direction.
The proof uses filling disks where the zj are the centres. Since (7) implies (1)
the detected Hayman direction is again a Julia direction. The condition (7) has
been studied in its own right. In particular it was proved by Toppila [11] that
there exists a sequence for which (7) holds if f has a Nevanlinna deficient value.
The following corollary seems not to have been stated before, but it follows
immediately from [11] and Theorem 4.
b that has a
Corollary 5. Every transcendental meromorphic function f : C → C
Nevanlinna deficient value has a Julia direction that is a Hayman direction.
60
A. Sauer
CMFT
At this point it is worth noting that a Valiron deficient value already implies
the existence of a Julia direction (see [16]). Toppila [11] also considered Valiron
deficiencies. His result for this case implies, again with Theorem 4:
b that has a
Corollary 6. Every transcendental meromorphic function f : C → C
Valiron deficient value and satisfies
T (r, f )
→∞
(log r)2
has a Julia direction that is a Hayman direction.
Unlike the case of Julia directions, where (1) implies the proper growth condition,
it seems that (7) does not imply (5); if
log |z|
#
(8)
f (z) = O
|z|
then standard estimations show A(r, f ) = O((log r)3 ) and T (r, f ) = O((log r)4 ).
On the other hand, we do not have an example with T (r, f ) ∼ (log r)4 and (8).
Now we will consider directions that are not Hayman directions and derive results
for the spherical derivative of the functions
fk (z) := z −k f (z)
with k ∈ N, that are analogous to (2). For this we need the normal family version
of Hayman’s Theorem (see [9, Cor. 4.5.9]).
Theorem 7. Let F be a family of meromorphic functions on a domain G such
that there exist a ∈ C, b ∈ C \ {0} and k ∈ N with f (z) 6= a and f (k) (z) 6= b for
all f ∈ F and z ∈ G. Then F is a normal family.
From this we can prove the following lemma.
b be meromorphic such that f
e be a sector and let f : ∆
e →C
Lemma 8. Let ∆
(k)
omits a finite value and f omits a non-zero finite value for some k ∈ N. Then
the function fk (z) := z −k f (z) satisfies fk# (z) = O(1/|z|) in any smaller sector
e
∆ ⊂ ∆.
e is strictly less than π.
Proof. We can assume that the opening angle δ of ∆
e First we will show
Further let eiα be the direction of the angle bisector of ∆.
that every sequence gxn in the family F of all functions of the form
b
gx : D → C,
gx (w) := x−k f (x(eiα + sw))
where s := sin(δ/2) and x > 1 possesses a subsequence that converges locally
uniformly in D. If xn has a finite accumulation point x
e then choose a subsequence
xnk → x
e and it follows that gxnk → gxe locally uniformly because of continuity.
Suppose xn → ∞. First we assume that f omits 0. Then gxn omits 0 and
gx(k)
(w) = sk f (k) (x(eiα + sw))
n
9 (2009), No. 1
How to Detect Hayman Directions
61
omits sk b 6= 0 where b 6= 0 is the value omitted by f (k) . By Theorem 7 it follows
that {gxn } is a normal family and thus possesses a convergent subsequence. If f
omits a 6= 0 then consider
iα
−k
gexn (w) := x−k
n [f (xn (e + sw)) − a] = gxn (w) − xn a.
(k)
Then gexn omits 0 and gexn omits sk b 6= 0. Again it follows that {e
gxn } is normal.
−k
Because xn a → 0 independently of w the same holds for {gxn }. From this
and Marty’s Theorem we get for |w| ≤ r < 1 the existence of a constant Kr
independent of x > 1 such that
(9)
x−k+1 |f 0 (x(eiα + sw))|
≤ Kr .
1 + |x−k f (x(eiα + sw))|2
Note that the left hand side of (9) is equal to gx# (w)/s.
Since fk0 (z) = −kz −k−1 f (z) + z −k f 0 (z) we get
fk# (z) ≤ k|z|−1
−k+1 0
|z −k f (z)|
|f (z)|
−1 |z|
+
|z|
−k
2
−k
1 + |z f (z)|
1 + |z f (z)|2
The first term is O(1/|z|) since c/(1 + c2 ) is bounded for c ≥ 0. It remains
to be shown that |z|−k+1 |f 0 (z)|/(1 + |z −k f (z)|2 ) is bounded in ∆. For ∆ there
exists r with 0 < r < 1 such that every z ∈ ∆ with |z| > 1 can be written as
z = x(eiα + sw) =: xζw with x > 1 and |w| ≤ r. With (9) it follows
|z|−k+1 |f 0 (z)|
|ζw |−k+1 x−k+1 |f 0 (xζw )|
=
1 + |z −k f (z)|2
1 + |ζw |−2k |x−k f (xζw )|2
x−k+1 |f 0 (xζw )| |ζw |−k+1 (1 + |x−k f (xζw )|2 )
=
1 + |x−k f (xζw )|2 1 + |ζw |−2k |x−k f (xζw )|2
|ζw |−k+1 (1 + |x−k f (xζw )|2 )
≤ Kr
1 + |ζw |−2k |x−k f (xζw )|2
Since δ < π we have 0 < s < 1 and obviously 1 − s < |ζw | < 1 + s. Set
C := (1 − s)−k+1 (1 + s)2k . Then it is easy to check that
|ζw |−k+1 (1 + |x−k f (xζw )|2 )
≤ C,
1 + |ζw |−2k |x−k f (xζw )|2
which proves the lemma.
We deduce the following theorem.
b be a meromorphic function. If
Theorem 9. Let f : C → C
lim sup |z|fk# (z) = ∞
z→∞
for all k ∈ N in every sector around some direction H, then H is a Hayman
direction.
62
A. Sauer
CMFT
To illustrate the usefulness of this theorem, we consider Ostrowski’s example
F (z) given in (3). This function is a Julia exceptional function, which implies
that F satisfies (2). From Marty’s Theorem the condition (2) is easily seen
to be equivalent to the normality of every sequence F (σn z) in C \ {0} with
arbitrary σn → ∞. Similarly, it follows from Marty’s Theorem that if for some k
the sequence Fk (σn z) = F (σn z)/(σn z)k is not normal in a sector for a positive
sequence σn → ∞, then |z|Fk# (z) is unbounded in that sector. We set σn = q n
and consider gn (z) := F (q n z) in C \ {0}. Then gn (1) = 0 and gn (−1) = ∞.
It follows that every limit g of the normal sequence gn is non-constant with
g(1) = 0 and g(−1) = ∞. Given any sector ∆ around the negative real axis,
one can find a neighbourhood U of z = −1 contained in ∆. For each k, the
sequence Fk (q n z) = gn (z)/(q n z)k is not normal in U , since a limit must be ∞
at z = −1 and 0 in a punctured neighbourhood of z = −1. This follows from
the fact that we may assume that gn has a non-constant limit. It follows that
Theorem 9 can be applied, so for Ostrowski’s example the negative real axis is
a Hayman direction that is not a Julia direction. To the best of our knowledge
this is the first example of this type.
From Lemma 8 we also get with a standard compactness argument:
b that does not have
Theorem 10. For every meromorphic function f : C → C
a Hayman direction there exists a covering of C by a finite number of sectors
∆1 , . . . , ∆n such that for every ∆l there exists kl ∈ N with
1
#
(10)
fkl (z) = O
|z|
in ∆l .
We propose to call a meromorphic function with such a sectorial covering of C a
Hayman exceptional function.
It is tempting to hope that Hayman exceptional functions satisfy (4), especially
since this would solve a problem that has remained unsolved for some time
now (see [1, Problem 1.31] and [15, Problem p. 121]). But this is not true:
g(z) := z · G(z), with G being Rossi’s example (6), is a Hayman exceptional
function with T (r, g) ∼ (log r)3 /3. The filling disks of G accumulate to the
imaginary axis, so, with the notation of Lemma 8, |z|g1# (z) is bounded in any
sector avoiding the imaginary axis. Around the imaginary axis g2 is bounded
and hence |z|g2# (z) is also bounded. (For these facts we again refer to [7, 2].)
On the other hand, a Hayman exceptional function cannot grow faster than the
above example.
b
Theorem 11. For every Hayman exceptional function f : C → C,
T (r, f ) = O((log r)3 ).
9 (2009), No. 1
How to Detect Hayman Directions
63
Proof. We use a lemma in [12, p. 277] in each of the sectors as described in
Theorem 10. With the notation of this lemma we set w := fkl , g1 (z) := z kl ,
g2 = g3 := 0 and g4 := 1, so that
g1 (z)w(z) + g2 (z)
.
f (z) =
g3 (z)w(z) + g4 (z)
From (10) it follows for the function denoted in the lemma by S(·, r; ∆) that
S(w, r; ∆l ) = O(log r). From T (r, g1 ) = O(log r) and hence
Z 128r
T (t, g1 )
dt = O((log r)2 )
t
1
we obtain (by decreasing the ∆l slightly, such that they still cover the whole
plane)
S(f, r; ∆l ) = O(log(64r)) + O((log r)2 ) = O((log r)2 ).
P
From A(r, f ) ≤ nl=1 S(f, r; ∆l ) + O(1), where A(r, f ) again denotes the unintegrated Ahlfors–Shimizu characteristic, we get A(r, f ) = O((log r)2 ). This proves
the theorem.
This reproves Yang Lo’s Theorem (at least the part on Hayman directions), but
not more and also in a rather indirect manner. Maybe further research on Hayman exceptional functions will show whether Yang Lo’s growth condition (5) is
sharp. Note that so far there exists no example of a transcendental meromorphic
function without a Hayman direction.
3. Julia directions of derivatives
We now consider a problem that has certain similarities with the study of Hayman
directions, namely the existence of Julia directions for f = f (0) and its derivatives
f (k) . We have already obtained some results in this direction in [8].
It seems that up to now it is not even known, whether there exist transcendental
meromorphic functions such that f and all its derivatives f (k) have no Julia
direction. With the results from the last section we can prove that there exist
no such functions.
b be a transcendental meromorphic function. Then
Theorem 12. Let f : C → C
there exists n in N∪{0} such that f (k) possesses a Julia direction for every k ≥ n.
Proof. Suppose that f is a Julia exceptional function. As already mentioned,
Ostrowski [6, Satz 5] gave a characterization of Julia exceptional functions, and
one condition [6, Satz 5, Part a)] a Julia exceptional function has to satisfy is
|n(r, f, 0)−n(r, f, ∞)| = O(1). It follows that N (r, f, 0)−N (r, f, ∞) ≤ c log r for
some c ≥ 0. Here n and N denote the usual counting functions from Nevanlinna
theory. Choose any k with k > c. If f (k) has no Julia direction, then f (k) omits
a finite non-zero value around every direction and we can apply Lemma 8 to f
64
A. Sauer
CMFT
and f (k) in a finite number of sectors that cover the whole plane. It turns out
that fk (z) = z −k f (z) has to be a Julia exceptional function with
N (r, fk , 0) − N (r, fk , ∞) = N (r, f, 0) − N (r, f, ∞) − k log r
≤ (c − k) log r → −∞.
But this contradicts another of Ostrovski’s conditions [6, Satz 5, Part c)]. Thus
f (k) possesses a Julia direction.
If none of the f (k) is Julia exceptional, we have nothing to prove. If there is one
such f (k) we can apply the above reasoning to it, rather than to f .
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Andreas Sauer
E-mail: [email protected]
Address: FH Dortmund, Sonnenstr. 96, 44047 Dortmund, Germany.