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volume 5, issue 2, article 35, 2004.

Received 12 October, 2003;

accepted 01 January, 2004.

Communicated by:H.M. Srivastava

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Journal of Inequalities in Pure and Applied Mathematics

SOME NORMALITY CRITERIA

INDRAJIT LAHIRI AND SHYAMALI DEWAN

Department of Mathematics University of Kalyani West Bengal 741235, India.

EMail:indrajit@cal2.vsnl.net.in Department of Mathematics University of Kalyani West Bengal 741235, India.

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2000Victoria University ISSN (electronic): 1443-5756 144-03

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Some Normality Criteria Indrajit Lahiri and Shyamali Dewan

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J. Ineq. Pure and Appl. Math. 5(2) Art. 35, 2004

http://jipam.vu.edu.au

Abstract

In the paper we prove some sufficient conditions for a family of meromorphic functions to be normal in a domain.

2000 Mathematics Subject Classification:Primary 30D45, Secondary 30D35.

Key words: Meromorphic function, Normality.

1. Introduction and Results

LetCbe the open complex plane andD ⊂ Cbe a domain. A familyF of meromorphic functions defined in Dis said to be normal, in the sense of Montel, if for any sequencef_n∈ F there exists a subsequencef_n_j such thatf_n_j converges spherically, locally and uniformly inDto a meromorphic function or∞.

F is said to be normal at a pointz₀ ∈ D if there exists a neighbourhood of z₀in whichF is normal. It is well known thatF is normal inDif and only if it is normal at every point ofD.

It is an interesting problem to find out criteria for normality of a family of analytic or meromorphic functions. In recent years this problem attracted the attention of a number of researchers worldwide.

In 1969 D. Drasin [5] proved the following normality criterion.

Theorem A. LetFbe a family of analytic functions in a domainDanda(6= 0), bbe two finite numbers. If for everyf ∈ F,f⁰−afⁿ−bhas no zero thenF is normal, wheren(≥3)is an integer.

Chen-Fang [2] and Ye [21] independently proved that TheoremAalso holds for n = 2. A number of authors {cf. [3, 11, 12, 13, 16, 24]} extended Theo- rem Ato a family of meromorphic functions in a domain. Their results can be combined in the following theorem.

Theorem B. Let F be a family of meromorphic functions in a domain D and a(6= 0),b be two finite numbers. If for everyf ∈ F, f⁰ −afⁿ−b has no zero thenF is normal, wheren(≥3)is an integer.

Li [12], Li [13] and Langley [11] proved Theorem Bfor n ≥ 5, Pang [16]

proved for n = 4 and Chen-Fang [3], Zalcman [24] proved for n = 3. Fang-

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Yuan [6] showed that TheoremBdoes not, in general, hold forn = 2. For the casen= 2they [6] proved the following result.

Theorem C. Let F be a family of meromorphic functions in a domain D and a(6= 0),bbe two finite numbers. Iff⁰−af²−bhas no zero andf has no simple and double pole for everyf ∈ F thenF is normal.

Fang-Yuan [6] mentioned the following example from which it appears that the condition for eachf ∈ F not to have any simple and double pole is neces- sary for TheoremC.

Example 1.1. Letfn(z) = nz(z√

n−1)⁻² for n = 1,2, . . .andD : |z| < 1.

Then eachf_nhas only a double pole and a simple zero. Alsof_n⁰+f_n² =n(z√ n−

1)⁻⁴ 6= 0. Sincef_n^#(0) = n→ ∞asn→ ∞, it follows from Marty’s criterion that{f_n}is not normal inD.

However, the following example suggests that the restriction on the poles of f ∈ F may be relaxed at the cost of some restriction imposed on the zeros of f ∈ F.

Example 1.2. Let fn(z) =nz⁻² forn = 3,4, . . .andD : |z| <1. Then each f_n has only a double pole and no simple zero. Also we see that f_n⁰ +f_n² = n(n−2z)z⁻⁴ 6= 0inD. Since

f_n^#(z) = 2n|z|

|z|²+n² ≤ 2 n <1

inD, it follows from Marty’s criterion that the family{f_n}is normal inD.

Now we state the first theorem of the paper.

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Theorem 1.1. LetF be a family of meromorphic functions in a domainDsuch that nof ∈ F has any simple zero and simple pole. Let

E_f ={z :z ∈ D and f⁰(z)−af²(z) =b}, wherea(6= 0),bare two finite numbers.

If there exists a positive numberM such that for everyf ∈ F, |f(z)| ≤ M wheneverz ∈Ef, thenF is normal.

The following examples together with Example1.1show that the condition of Theorem1.1on the zeros and poles are necessary.

Example 1.3. Letf_n(z) =ntannzforn = 1,2, . . .andD :|z| < π. Thenf_n has only simple zeros and simple poles. Also we see that f_n⁰ −f_n² = n² 6= 0.

Sincef_n^#(0) = n² → ∞asn→ ∞, by Marty’s criterion the family{f_n}is not normal.

Example 1.4. Let f_n(z) = (1 +e^2nz)⁻¹ for n = 1,2, . . . and D : |z| < 1.

Thenf_nhas no simple zero and no multiple pole. Also we see thatf_n⁰ +f_n² 6= 1.

Sincef_n^#(0) = ²ⁿ₃ → ∞asn→ ∞, by Marty’s criterion the family{fn}is not normal.

Drasin [18, p. 130] also proved the following normality criterion which involves differential polynomials.

Theorem D. LetF be a family of analytic functions in a domainDanda₀, a₁, . . . , ak−1be finite constants, wherekis a positive integer. Let

H(f) =f^(k)+ak−1f^(k−1)+. . .+a₁f⁽¹⁾+a₀f.

If for everyf ∈ F

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(i) f has no zero,

(ii) H(f)−1has no zero of multiplicity less thank+ 2, thenF is normal.

Recently Fang-Yuan [6] proved that TheoremDremains valid even ifH(f)−

1 has only multiple zeros for every f ∈ F. In the next theorem we extend TheoremDto a family of meromorphic functions which also improves a result of Fang-Yuan [6].

Theorem 1.2. LetF be a family of meromorphic functions in a domainDand H(f) =f^(k)+ak−1f^(k−1)+. . .+a₁f⁽¹⁾+a₀f,

wherea0, a1, . . . , ak−1are finite constants andkis a positive integer.

Let

E_f ={z :z ∈ Dandzis a simple zero ofH(f)−1}.

(i) f has no pole of multiplicity less than3 +k, (ii) f has no zero,

(iii) there exists a positive constantM such that|f(z)| ≥M wheneverz ∈E_f, thenF is normal.

The following examples show that conditions (ii) and (iii) of Theorem 1.2 are necessary, leaving the question of necessity of the condition (i) as open.

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Example 1.5. Letf_n(z) = nzforn = 2,3, . . .,D:|z|<1,H(f) =f⁰−f and M = ¹₂. Then eachfnhas a zero atz = 0andEfn ={1− ¹_n}forn= 2,3, . . ..

So |f(1− _n¹)| = n −1 ≥ M for n = 2,3, . . .. Since f_n^#(0) = n → ∞as n → ∞, by Marty’s criterion the family{f_n}is not normal inD.

Example 1.6. Letfn(z) =e^nzforn = 2,3, . . .,D:|z|<1andH(f) =f⁰−f. Then each f_n has no zero andE_f_n = {z : z ∈ D and (n−1)e^nz = 1}for n = 2,3, . . .. Also we see that for z ∈ E_f_n, |f_n(z)| = _n−1¹ → 0as n → ∞.

Sincef_n^#(0) = ⁿ₂ → ∞asn → ∞, by Marty’s criterion the family{fn}is not normal inD.

In connection to TheoremAChen-Fang [3] proposed the following conjecture:

Conjecture 1. Let F be a family of meromorphic functions in a domainD. If for every function f ∈ F, f^(k)−afⁿ−b has no zero inD thenF is normal, wherea(6= 0),bare two finite numbers andk, n(≥k+ 2)are positive integers.

In response to this conjecture Xu [23] proved the following result.

Theorem E. Let F be a family of meromorphic functions in a domain D and a(6= 0), b be two finite constants. If k and n are positive integers such that n ≥k+ 2and for everyf ∈ F

(i) f^(k)−afⁿ−bhas no zero, (ii) f has no simple pole, thenF is normal.

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The condition (ii) of Theorem E can be dropped if we choose n ≥ k + 4 (cf. [15,17]). Also some improvement of TheoremEcan be found in [22]. In the next theorem we investigate the situation when the power offis negative in condition (i) of TheoremE.

Theorem 1.3. LetF be a family of meromorphic functions in a domainD and a(6= 0),bbe two finite numbers. Suppose thatE_f ={z :z ∈ D and f^(k)(z) + af⁻ⁿ(z) = b}, wherek, n(≥k)are positive integers.

(i) f has no zero of multiplicity less thank,

(ii) there exists a positive number M such that for everyf ∈ F, |f(z)| ≥ M wheneverz ∈E_f,

thenF is normal.

Following examples show that the conditions of Theorem1.3are necessary.

Example 1.7. Letf_p(z) = pz² forp = 1,2, . . .andD : |z| < 1, n = k = 3, a = 1,b= 0. Thenfphas only a double zero andEfp =∅. Sincefp(0) = 0and forz 6= 0,f_p(z)→ ∞asp→ ∞, it follows that the family{f_p}is not normal.

Example 1.8. Let fp(z) = pz for p = 1,2, . . . and D : |z| < 1, n = k = 1.

Thenf_phas simple zero at the origin and for any two finite numbersa(6= 0),b, E_f_p ={a/p(b−p)}so that|f_p(z)| → 0asp → ∞wheneverz ∈ E_f_p. Since f_p^#(0) =p→ ∞asp→ ∞, by Marty’s criterion the family{fp}is not normal.

For the standard definitions and notations of the value distribution theory we refer to [8,18].

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2. Lemmas

In this section we present some lemmas which will be needed in the sequel.

Lemma 2.1. [1] Let f be a transcendental meromorphic function of finite or- der in C. Iff has no simple zero then f⁰ assumes every non-zero finite value infinitely often.

Lemma 2.2. [10] Let f be a nonconstant rational function in C having no simple zero and simple pole. Thenf⁰assumes every non-zero finite value.

The following lemma can be proved in the line of [9].

Lemma 2.3. Letf be a meromorphic function inCsuch thatf^(k) 6≡0. Suppose thatψ = fⁿf^(k), wherek, nare positive integers. Ifn > k = 2orn ≥ k ≥ 3 then

1− 1 +k

n+k − n(1 +k) (n+k)(n+k+ 1)

T(r, ψ)≤N(r, a;ψ) +S(r, ψ), wherea(6= 0,∞)is a constant.

Lemma 2.4. [19] Let f be a transcendental meromorphic function in C and ψ =fⁿf⁽²⁾, wheren(≥2)is an integer. Then

lim sup

r→∞

N(r, a;ψ) T(r, ψ) >0, wherea(6= 0,∞)is a constant.

The following lemma is a combination of the results of [3,7,14].

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Lemma 2.5. Letf be a transcendental meromorphic function inC. Thenfⁿf⁰ assumes every non-zero finite value infinitely often, wheren(≥1)is an integer.

Lemma 2.6. Letf be a non-constant rational function inC. Thenfⁿf⁰assumes every non-zero finite value.

Proof. Letg =fⁿ⁺¹/(n+ 1). Theng is a nonconstant rational function having no simple zero and simple pole. So by Lemma 2.2 g⁰ = fⁿf⁰ assumes every non-zero finite value. This proves the lemma.

Lemma 2.7. Let f be a rational function inC such thatf⁽²⁾ 6≡ 0. Thenψ = f²f⁽²⁾assumes every non-zero finite value.

Proof. Let f = p/q, wherep, q are polynomials of degree m, n respectively andp,qhave no common factor.

Letabe a non-zero finite number. We now consider the following cases.

Case 1. Let m = n. Thenf = α+p₁/q, where α is a constant and p₁ is a polynomial of degreem₁ < n.

Now

f⁰ = p⁰₁q−p₁q⁰ q² = p₂

q₂, say,

wherep₂ andq₂are polynomials of degreem₂ =m₁+n−1andn₂ = 2n. Also we note thatm₂ < n₂. Hence

f⁰⁰ = p⁰₂q₂−p₂q⁰₂ q₂² = p₃

q3

, say,

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wherep₃ andq₃ are polynomials of degreem₃ =m₂+n₂−1 =m₁ + 3n−2 andn3 = 2n2 = 4n. Also we see thatm3 < n3.

Letψ =f²f⁽²⁾ =P/Q. ThenP,Qare polynomials of degree2m+m₃ and 2n+n₃respectively and2m+m₃ <2n+n₃. Thereforeψ is nonconstant.

Nowψ−a= (P−aQ)/Qand the degree ofP−aQis equal to the degree of Q. Ifψ−ahas no zero thenP−aQandQshare0CM (counting multiplicites) and soP −aQ ≡ AQ, whereAis a constant. Thereforeψ = A−a, which is impossible. Soψ−amust have some zero.

Case 2. Letm=n+ 1. Then

f =αz+β+ p₁ q ,

whereα,β are constants andp₁is a polynomial of degreem₁ < n.

Nowf⁰⁰ = p3/q3, wherep3 and q3 are polynomials of degree m3 = m1 + 3n−2andn₃ = 4nrespectively andm₃ < n₃.

If ψ = P/Q thenP, Qare polynomials of degree 2m+m₃ and 2n+n₃ respectively. We see that 2m+m₃ = 5n +m₁ < 6n = 2n+n₃ and so ψ is nonconstant. Therefore as Case1ψ−amust have some zero.

Case 3. Letm6=n, n+ 1. Then

f⁰ = pq⁰−p⁰q q² = p₄

q₄, say,

wherep₄,q₄are polynomials of degreem₄ =m+n−1andn₄ = 2n. Also we note thatm₄ 6=n₄.

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Hence

f⁰⁰ = p⁰₄q₄−p₄q⁰₄ q₄² = p₅

q₅, say,

where p₅,q₅ are polynmials of degreem₅ =m₄+n₄−1 = m+ 3n−2and n5 = 2n4 = 4n.

If ψ = P/Q thenP, Qare polynomials of degree 2m+m₅ and 2n+n₅ respectively. Clearly 2m+m₅ 6= 2n+n₅ because otherwise m = n+ 2/3, which is impossible. Soψis nonconstant. Also we see thatψ−a= (P−aQ)/Q, where the degree ofP −aQis not less than that ofQ. Ifψ−ahas no zero then as per Case 1ψ becomes a constant, which is impossible. Soψ −amust have some zero. This proves the lemma.

Lemma 2.8. Let f be a meromorphic function in C such that f^(k) 6≡ 0 and a(6= 0) be a finite constant. Then f^(k)+af⁻ⁿ must have some zero, where k andn(≥k)are positive integers.

Proof. First we assume that k = 1. Then by Lemmas2.5 and 2.6 we see that fⁿf⁰+amust have some zero. Since a zero offⁿf⁰ +ais not a pole or a zero off, it follows that a zero offⁿf⁰+ais a zero off⁰+af⁻ⁿ.

Now we assume thatk = 2. Then by Lemmas2.3, 2.4 and2.7 we see that fⁿf⁽²⁾+amust have some zero. As the preceding paragraph a zero offⁿf⁽²⁾+a is a zero off⁽²⁾+af⁻ⁿ.

Finally we assume that k ≥ 3. Then by Lemma 2.3 fⁿf^(k)+a must have some zero. Since a zero of fⁿf^(k)+ais a zero of f^(k)+af⁻ⁿ, the lemma is proved.

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Lemma 2.9. Letf be a nonconstant meromorphic function inCsuch thatfhas no zero and has no pole of multiplicity less than3 +k. Thenf^(k)−1must have some simple zero, wherekis a positive integer.

Proof. Since N(r, f^(k)) = N(r, f) + kN(r, f) and m(r, f^(k)) ≤ m(r, f) + S(r, f), we get

T(r, f^(k))≤T(r, f) +kN(r, f) +S(r, f)

≤T(r, f) + k

3 +kN(r, f) +S(r, f)

≤ 3 + 2k

3 +k T(r, f) +S(r, f).

Sincefhas no zero and no pole of multiplicity less than3+k, we get by Milloux inequality ([8, p. 57])

T(r, f)≤N(r, f) +N(r,1;f^(k)) +S(r, f)

≤ 1

3 +kT(r, f) +N(r,1;f^(k)) +S(r, f).

If possible, suppose thatf^(k)−1has no simple zero. Then we get from above T(r, f)≤ 1

3 +kT(r, f) + 1

2N(r,1;f^(k)) +S(r, f)

≤ 1

3 +k + 3 + 2k 2(3 +k)

T(r, f) +S(r, f)

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and so

1

2(3 +k)T(r, f)≤S(r, f), a contradiction. This proves the lemma.

Lemma 2.10. [4,20] LetF be a family of meromorphic functions in a domain D and let the zeros of f be of multiplicity not less thank ( a positive integer) for eachf ∈ F. IfF is not normal atz₀ ∈ Dthen for0 ≤α < kthere exist a sequence of complex numbers z_j → z₀, a sequence of functionsf_j ∈ F, and a sequence of positive numbersρ_j →0such that

g_j(ζ) = ρ^−α_j f_j(z_j+ρ_jζ)

converges spherically and locally uniformly to a nonconstant meromorphic func- tiong(ζ)inC. Moreover the order ofg is not greater than two and the zeros of g are of multiplicity not less thank.

Note 1. If each f ∈ F has no zero theng also has no zero and in this case we can chooseαto be any finite real number.

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3. Proofs of the Theorems

In this section we discuss the proofs of the theorems.

Proof of Theorem1.1. If possible suppose thatF is not normal atz₀ ∈ D. Then F₁ ={1/f :f ∈ F }is not normal atz₀ ∈ D. Letα= 1. Then by Lemma2.10 there exist a sequence of functions f_j ∈ F, a sequence of complex numbers z_j →z₀ and a sequence of positive numbersρ_j →0such that

g_j(ζ) =ρ⁻¹_j f_j⁻¹(z_j +ρ_jζ)

converges spherically and locally uniformly to a nonconstant meromorphic fuc- ntion g(ζ) inC. Also the order ofg does not exceed two and g has no simple zero. Again by Hurwitz’s theoremg has no simple pole.

By Lemmas2.1and2.2we see that there existsζ₀ ∈Csuch that

(3.1) g⁰(ζ₀) +a= 0.

Since ζ₀ is not a pole of g, it follows that g_j(ζ) converges uniformly to g(ζ) in some neighbourhood ofζ₀. We also see that _g⁻¹2(ζ){g⁰(ζ) +a}is the uniform limit ofρ²_j{f_j⁰ −af_j²−b}in some neighbourhood ofζ0.

In view of (3.1) and Hurwitz’s theorem there exists a sequenceζ_j →ζ₀ such thatf_j⁰(ζ_j)−af_j²(ζ_j)−b= 0. So by the given condition

|g_j(ζ_j)|= 1

ρ_j · 1

|f_j(z_j +ρ_jζ_j)| ≥ 1 ρ_jM.

Sinceζ₀is not a pole ofg, there exists a positive numberKsuch that in some neighbourhood ofζ0we get|g(ζ)| ≤K.

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Sinceg_j(ζ) converges uniformly to g(ζ)in some neighbourhood of ζ₀, we get for all large values ofj and for allζin that neighbourhood ofζ0

|g_j(ζ)−g(ζ)|<1.

Sinceζ_j →ζ, we get for all large values ofj

K ≥ |g(ζ_j)| ≥ |g_j(ζ_j)| − |g(ζ_j)−g_j(ζ_j)|> 1 ρ_jM −1, which is a contradiction. This proves the theorem.

Proof of Theorem1.2. Let α = k. If possible suppose thatF is not normal at z₀ ∈ D. Then by Lemma2.10 and Note1there exists a sequence of functions f_j ∈ F, a sequence of complex numbers z_j → z₀ and a sequence of positive numbersρj →0such that

gj(ζ) =ρ^−k_j fj(zj +ρjζ)

converges spherically and locally uniformly to a nonconstant meromorphic func- tiong(ζ)inC. Now by conditions (i) and (ii) and by Hurwitz’s theorem we see thatg(ζ)has no zero and has no pole of multiplicity less than3 +k.

Now by Lemma2.9g^(k)(ζ)−1has a simple zero at a pointζ₀ ∈C. Sinceζ₀ is not a pole of g(ζ), in some neighbourhood of ζ₀,g_j(ζ)converges uniformly tog(ζ).

Since g_j^(k)(ζ)−1 +

k−1

X

i=0

a_iρ^k−i_j g⁽ⁱ⁾_j (ζ) = f_j^(k)(z_j+ρ_jζ) +

k−1

X

i=0

a_if_j⁽ⁱ⁾(z_j+ρ_jζ)−1

=H(f_j(z_j+ρ_jζ))−1

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andPk−1

i=0 a_iρ^k−i_j g_j⁽ⁱ⁾(ζ)converges uniformly to zero in some neighbourhood of ζ₀, it follows thatg^(k)(ζ)−1is the uniform limit ofH(f_j(z_j +ρ_jζ))−1.

Sinceζ₀ is a simple zero ofg^(k)(ζ)−1, by Hurwitz’s theorem there exists a sequenceζ_j → ζ₀ such thatζ_j is a simple zero ofH(f_j(z_j +ρ_jζ))−1. So by the given condition|f_j(z_j+ρ_jζ_j)| ≥M for all large values ofj.

Hence for all large values ofj we get|gj(ζj)| ≥M/ρ^k_j and as the last part of the proof of Theorem1.1we arrive at a contradiction. This proves the theorem.

Proof of Theorem1.3. Let α = k/(1 +n) < 1. If possible suppose that F is not normal atz₀ ∈ D. Then by Lemma2.10there exist a sequence of functions f_j ∈ F, a sequence of complex numbers z_j → z₀ and a sequence of positive numbersρ_j →0such that

g_j(ζ) = ρ^−α_j f_j(z_j+ρ_jζ)

converges spherically and locally uniformly to a nonconstant meromorphic func- tiong(ζ)inC. Alsog has no zero of multiplicity less thank. Sog^(k) 6≡ 0and by Lemma2.8we get

(3.2) g^(k)(ζ₀) + a

gⁿ(ζ₀) = 0 for someζ₀ ∈C.

Clearlyζ0is neither a zero nor a pole ofg. So in some neighbourhood ofζ0, g_j(ζ)converges uniformly tog(ζ).

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Now in some neighbourhood ofζ₀we see thatg^(k)(ζ) +ag⁻ⁿ(ζ)is the uniform limit of

g_j^(k)+ag_j⁻ⁿ(ζ)−ρ^nα_j b =ρ

nk 1+n

j

n

f_j^(k)(z_j +ρ_jζ) +af_j⁻ⁿ(z_j+ρ_jζ)−bo . By (3.2) and Hurwitz’s theorem there exists a sequence ζ_j → ζ₀ such that for all large values ofj

f_j^(k)(zj +ρjζj) +af_j⁻ⁿ(zj+ρjζj) =b.

Therefore for all large values ofjit follows from the given condition|g_j(ζ_j)| ≥ M/ρ^α_j and as in the last part of the proof of Theorem1.1we arrive at a contradiction. This proves the theorem.

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[10] I. LAHIRI, A simple normality criterion leading to a counterexample to the converse of Bloch principle, New Zealand J. Math., (to appear).

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