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NEW ZEALAND JOURNAL OF MATHEMATICS
Volume 34 (2005), 61–65
A SIMPLE NORMALITY CRITERION LEADING TO A
COUNTEREXAMPLE TO THE CONVERSE OF THE BLOCH
PRINCIPLE
Indrajit Lahiri
(Received August 2003)
Abstract. Proving a normality criterion for a family of meromorphic functions
we exhibit a counterexample to the general converse of Bloch principle.
1. Introduction
A well–known heuristic principle in the theory of functions is the Bloch principle.
Robinson [3] noted the Bloch principle as one of the twelve mathematical problems
requiring further consideration. The Bloch principle ([7]; p.101 [5]) is the hypothesis
that a family of analytic (meromorphic) functions which have a common property
P in a domain D will in general be a normal family if P reduces an analytic
(meromorphic) function in the open complex plane C to a constant. Unfortunately
the Bloch principle is not universally true. Some counterexamples can be found in
[4].
The converse of the Bloch principle can be stated as follows (p. 106 [5]):
If any family of meromorphic functions satisfying a property P in an arbitrary
domain is necessarily normal, then a function that is meromorphic in C and possesses the property P reduces to a constant.
Though the converse of the Bloch principle holds under some additional conditions, the general converse of the Bloch principle is not true (cf. pp. 107–108 [5]).
So far the author knows the counterexamples as mentioned in [5] are the only two
available in the literature.
The purpose of the paper is to prove a simple normality criterion for a family of
meromorphic functions and use it to provide another counterexample to the general
converse of the Bloch principle.
2. Lemmas
In this section we present some lemmas which will be required in the sequel.
Lemma 2.1. [1] Let f be a transcendental meromorphic function in C and
a(6= 0, ∞) be a constant. Then f f 0 + a has infinitely many zeros.
Lemma 2.2. Let g be a non–constant rational function having no simple zero and
no simple pole. Then for any constant a(6= 0, ∞) g 0 + a must have some zero.
1991 Mathematics Subject Classification Primary 30D45, Secondary 30D35.
Key words and phrases: Meromorphic functions, Normal families, Bloch principle.
62
I. LAHIRI
Proof. If g is a polynomial then the degree of g is at least two and so g 0 is a
non-constant polynomial. Hence g 0 + a has at least one zero.
Let g = p/q, where p, q are polynomials of degree m and n(≥ 1) respectively
and p, q have no common factor.
If possible we suppose that g 0 + a has no zero. Now we consider the following
cases.
Case 2.3. Let m < n + 1. Then
R
p0 q − pq 0 + aq 2
= ,
say.
g0 + a =
q2
S
Then R, S are non–constant polynomials such that degree of R = degree of S. Since
g 0 + a has no zero, it follows that R and S share zeros (counting multiplicities). So
R = AS, where A is a constant. Therefore g 0 + a = A and so g = (A − a)z + B,
where B is a constant. This is impossible because g is non–constant and has no
simple zero.
Case 2.4. Let m > n + 1. Then
p1
,
q
where p1 and r are polynomials with respective degrees m1 and t(≥ 2) such that
m = t + n and m1 < n. So
g = r+
(r0 + a)q 2 + p01 q − p1 q 0
q2
R1
=
,
say.
S
Let p1 = am1 z m1 +· · ·+a1 z+a0 and q = bn z n +· · ·+b1 z+b0 , where am1 6= 0, bn 6=
0. Since the coefficient of the leading term of p01 q − p1 q 0 is (m1 − n)am1 bn 6= 0, the
degree of p01 q − p1 q 0 is m1 + n − 1. Since m1 + n − 1 < 2n − 1 < 2n + t − 1, we
see that degree of R1 ≥ degree of S. So, as in the Case 2.3 , g becomes a linear
polynomial, which is impossible.
g0 + a = r0 +
p01 q − p1 q 0
+a
q2
=
Case 2.5. Let m = n + 1. Then
p1
,
q
where α, β are constants and p1 is a polynomial of degree m1 < n.
Let a + α 6= 0. Then
g = αz + β +
(2.1)
p01 q − p1 q 0 + (a + α)q 2
R2
=
,
say.
q2
S
Since the degree of p01 q − p1 q 0 is m1 + n − 1 < 2n − 1 < 2n, it follows that degree
of R2 = degree of S. So, as in the Case 2.3 , g becomes a linear polynomial, which
is impossible.
Let a + α = 0. Then
p0 q − p1 q 0
g0 + a = 1 2
,
(2.2)
q
where the degree of p01 q − p1 q 0 is m1 + n − 1.
Now we consider the following two subcases.
g0 + a =
A SIMPLE NORMALITY CRITERION
63
Subcase 2.6. Let p01 q − p1 q 0 have no zero. Then m1 + n = 1 and so m1 = 0 and
n = 1 because m1 < n. So from (2.1) we see that
g = αz + β +
D
,
γz + δ
where γ, δ, D are constants.
This is impossible because g is non–constant and has no simple zero and no
simple pole.
Subcase 2.7. Let p01 q − p1 q 0 have some zero. Since g 0 + a has no zero, it follows
that all the factors of p01 q − p1 q 0 are factors of q 2 . Hence we can write
q 2 = (p01 q − p1 q 0 )q1 ,
where q1 is a polynomial of degree n + 1 − m1 (> 0).
From (2.2) we get
1
.
(2.3)
q1
Let α1 , α2 , . . . , αl be distinct zeros of q1 with respective multiplicities k1 , k2 , . . . , kl .
Since q1 is non–constant, we see that l ≥ 1.
Since α1 , α2 , . . . , αl are the only poles of g 0 with respective multiplicities k1 , k2 , . . .,
kl , it follows that α1 , α2 , . . ., αl are the only poles of g with respective multiplicities
k1 − 1, k2 − 1, . . ., kl − 1. Hence we get from (2.1) and (2.3)
g0 + a =
k1 + k2 + · · · + kl = n + l
and k1 + k2 + · · · + kl = n + 1 − m1 .
So m1 + l = 1. Since l ≥ 1, it follows that m1 = 0, l = 1 and k1 = n + 1. Therefore
from (2.1) we get
g
D
(γz + δ)n
(αz + β)(γz + δ)n + D
.
(γz + δ)n
= αz + β +
=
(2.4)
Since g is not a linear polynomial, it follows that D 6= 0. Also we see that z = −δ/γ
is the only pole of g.
Let Q = (αz + β)(γz + δ)n + D. Then Q0 = (γz + δ)n−1 [α(γ + n)z + αδ + nβ].
Since g has no simple zero, it follows that a zero of g is also a zero of Q0 and so a
zero of α(γ + n)z + αδ + nβ. So g and Q has only one double zero. Hence we get
n = 1. Therefore from (2.4) we get
D
,
γz + δ
which is impossible because g is non–constant and has no simple zero and no simple
pole. This proves the lemma.
g = αz + β +
Remark 2.8. The condition that g should not have any simple pole and simple
zero is necessary. For, if g = (z − 1)2 /z then g 0 − 1 = −1/z 2 has no zero and if
g = (z 3 − 1)/z 2 then g 0 − 1 = 2/z 3 has no zero.
Lemma 2.9. Let f be a non–constant meromorphic function in C. Then f f 0 + a
must have some zero for any constant a(6= 0, ∞).
64
I. LAHIRI
Proof. If f is transcendental meromorphic, the lemma follows from Lemma 2.1.
If f is non–constant and rational, we put g = f 2 /2. Then g is a non–constant
rational function having no simple zero and simple pole. Therefore by Lemma 2.2
f f 0 + a ≡ g 0 + a has some zero. This proves the lemma.
Lemma 2.10. {[2];see also [6]} Let F be a family of meromorphic functions defined
in a domain D. If F is not normal at a point z0 ∈ D then for −1 < α < 1 there
exist a sequence of functions fj ∈ F, a sequence of complex numbers zj → z0 and
a sequence of positive numbers ρj → 0 such that gj (ζ) = ρα
j fj (zj + ρj ζ) converges
spherically and locally unifomly to a non–constant meromorphic function g in C.
3. A Normality Criterion
In this section we prove a normality criterion for a class of meromorphic functions.
Theorem 3.1. Let F be a family of meromorphic functions defined in a domain
D and a(6= 0), b be two finite constants. Let
a
0
=b .
Ef = z : z ∈ D and f (z) +
f (z)
If there exists a positive number M such that for every f ∈ F, |f (z)| ≥ M whenever
z ∈ Ef then F is normal.
Proof. Let α = 1/2. If possible, suppose that F is not normal at z0 ∈ D. Then
by Lemma 2.10 there exist a sequence of functions fj ∈ F, a sequence of complex
numbers zj → z0 and a sequence of positive numbers ρj → 0 such that
gj (ζ) = ρ−α
j fj (zj + ρj ζ)
converges spherically and locally uniformly to a non–constant meromorphic function
g(ζ) in C.
Now by Lemma 2.9 it follows that
a
g 0 (ζ0 ) +
= 0
(3.1)
g(ζ0 )
for some ζ0 ∈ C.
Clearly ζ0 is neither a zero nor a pole of g. So in some neighbourhood of ζ0 gj (ζ)
converges uniformly to g(ζ).
Now in some neighbourhood of ζ0 we see that
a
g 0 (ζ) +
g(ζ)
is the uniform limit of
a
a
0
α
α
0
gj (ζ) +
− ρj b = ρj fj (zj + ρj ζ) +
−b .
gj (ζ)
fj (zj + ρj ζ)
In view of (3.1) we see by Hurwitz’s theorem that there exists a sequence {ζj }
converging to ζ0 such that for all large values of j
a
fj0 (zj + ρj ζj ) +
= b.
fj (zj + ρj ζj )
Therefore for all large values of j it follows from the given condition that
|fj (zj + ρj ζj )| ≥ M and so |gj (ζj )| ≥ ρ−α
j M.
A SIMPLE NORMALITY CRITERION
65
Since g is regular at ζ0 , in some neighbourhood of ζ0 we see that |g(ζ)| ≤ K for
some positive constant K. Again for all ζ in this neighbourhood of ζ0 and for all
large values of j we get |gj (ζ) − g(ζ)| < 1. So
K ≥ |g(ζj )| ≥ |gj (ζj )| − |gj (ζj ) − g(ζj )| ≥ ρ−α
j M −1
for all large values of j. This contradiction proves the theorem.
Corollary 3.2. Let F be a family of meromorphic functions defined in a domain
D and a(6= 0), b be two finite constants. If for every f ∈ F, f 0 + fa − b has no zero
in D then F is normal.
4. A Counter Example to the Converse of Bloch Principle
Let us say that a meromorphic function f possesses the property P if
1
f0 + − 1
f
has no zero. From the corollary we see that if every member of a family F of
meromorphic functions in any domain D possesses the property P then F is normal.
The following example shows that in contrast to the converse of the Bloch principle there exists a non–constant entire function which possesses the property P in
C.
Example 4.1. Let f = 1 + ez . Then
f0 +
1
e2z
−1 =
f
1 + ez
has no zero in C.
References
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A), 38 No. 7 (1995), 789–798.
2. X.C. Pang, Criteria for normality about differential polynomial , Chinese Sci.
Bull. 22 (1988), 1690–1693.
3. A. Robinson, Mathematical problems, J. Symbolic Logic, 38 (1973), 500–516.
4. L. Rubel, Four counterexamples to Bloch’s principle, Proc. Amer. Math. Soc.
98 (1986), 257–260.
5. J.L. Schiff, Normal Families, Springer–Verlag, 1993.
6. Y. Xu and X. Hua, Normality criteria of families of meromorphic functions,
Indian J. Pure Appl.Math. 31 No. 1 (2000), 61–68.
7. L. Zalcman, A heuristic principle in complex function theory, Amer. Math.
Monthly, 82 (1975), 813–817.
Indrajit Lahiri
Department of Mathematics
University of Kalyani
West Bengal 741235
INDIA
[email protected]