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ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 3 Issue 1(2011), Pages 14- 26.

GROWTH OF SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS IN THE UNIT DISC

(COMMUNICATED BY FIORALBA CAKONI)

BENHARRAT BELA¨IDI

Abstract. In this paper, we investigate the growth of solutions of higher order linear differential equations with analytic coefficients in the unit disc ∆ ={z∈C:|z|<1}. We obtain four results which are similar to those in the complex plane.

1. Introduction and statement of results

The growth of solutions of the complex linear differential equation

f(n)+an−1(z)f(n−1)+...+a1(z)f

0

+a0(z)f = 0 (1.1)

is well understood when the coefficients aj are polynomials [16, 17, 22, 27] or of finite (iterated) order of growth in the complex plane [7, 14, 15, 21]. In particular, Wittich [27] showed that all solutions of (1.1) are entire functions of finite order if and only if all coefficients are polynomials, and Gundersen, Steinbart and Wang [17] listed all possible orders of solutions of (1.1) in terms of the degrees of the polynomial coefficients.

Recently, there has been an increasing interest in studying the growth of ana-lytic solutions of linear differential equations in the unit disc ∆ by making use of Nevanlinna theory. The analysis of slowly growing solutions have been studied in [13, 19, 25]. Fast growth of solutions are considered by [8, 9, 11, 12, 19, 20].

In this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna’s theory on the complex plane and in the unit disc ∆ ={z∈C:|z|<1}(see [18, 19, 24, 26, 28, 29]).

We need to give some definitions and discussions. Firstly, let us give definition about the degree of small growth order of functions in ∆ as polynomials on the complex plane C. There are many types of definitions of small growth order of functions in ∆ (i.e., see [12, 13]).

1991Mathematics Subject Classification. 34M10, 30D35.

Key words and phrases. Linear differential equations; analytic solutions; unit disc.

c

2011 Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e. Submitted April 25, 2010. Published December 13, 2010.

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Definition 1.1. For a meromorphic function f in∆ let

D(f) := lim r→1−sup

T(r, f)

−log (1−r),

where T(r, f) is the characteristic function of Nevanlinna of f. If D(f)<∞, we say that f is of finite degree D(f) (or is non-admissible); if D(f) = ∞, we say thatf is of infinite degree (or is admissible). Iff is an analytic function in∆, and

DM(f) := lim r→1−sup

log+M(r, f)

−log (1−r) ,

in which M(r, f) = max

|z|=r|f(z)| is the maximum modulus function, then we say

that f is a function of finite degree DM(f) if DM(f) < ∞; otherwise, f is of

infinite degree.

Now, we give the definitions of iterated order and growth index to classify generally the functions of fast growth in ∆ as those in C (see [7, 21, 22]). Let us define inductively, forr∈[0,1), exp1r:=erand exp

p+1r:= exp exppr

, p∈N. We also define for allrsufficiently large in (0,1),log1r:= logrand logp+1r:= log logpr

, p ∈ N. Moreover, we denote by exp0r := r, log0r := r, log1r := exp1r and exp1r:= log1r.

Definition 1.2. [8, 11, 20] Let f be a meromorphic function in ∆. Then the iterated p−order off is defined by

ρp(f) = lim r→1−sup

log+pT(r, f)

−log (1−r) (p>1 is an integer),

where log+1 x = log+x = max{logx,0}, log+p+1x = log+log+p x. For p = 1, this notation is called order and for p= 2hyper order [19, 23]. If f is analytic in∆, then the iterated p−order off is defined by

ρM,p(f) = lim r→1−sup

log+p+1M(r, f)

−log (1−r) (p>1 is an integer).

Remark 1.3. It follows by M. Tsuji ([26], p. 205) that iff is an analytic function in∆, then we have the inequalities

ρ1(f)6ρM,1(f)6ρ1(f) + 1,

which are the best possible in the sense that there are analytic functions g and h such that ρM,1(g) =ρ1(g)andρM,1(h) =ρ1(h) + 1, see [13]. However, it follows

by Proposition 2.2.2 in [22]that ρM,p(f) =ρp(f)forp>2.

Definition 1.4. (see[11]) The growth index of the iterated order of a meromorphic function f(z)in ∆is defined by

i(f) =

0, iff is non-admissible,

min{j∈N:ρj(f)<+∞} if f is admissible, +∞, ifρj(f) = +∞for allj∈N.

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iM(f) =

0, iff is non-admissible,

min{j∈N:ρM,j(f)<+∞} if f is admissible, +∞, if ρM,j(f) = +∞for allj ∈N.

Remark 1.5. If ρp(f) <∞ or i(f) 6p, then we say that f is of finite iterated

p−order; if ρp(f) = ∞ or i(f) > p, then we say that f is of infinite iterated

p−order. In particular, we say that f is of finite order if ρ1(f)<∞or i(f)61; f is of infinite order ifρ1(f) =∞ ori(f)>1.

In [2], the author and Hamouda have proved the following result.

Theorem 1.6. [2] Let B0(z), ..., Bn−1(z) with B0(z) 6≡ 0 be entire functions.

Suppose that there exist a sequence of complex numbers (zj)jN with lim

j→+∞zj=∞

and three real numbersα, β andµ satisfying06β < α andµ >0 such that

|B0(zj)|>exp{α|zj| µ

}

and

|Bk(zj)|6exp{β|zj| µ

} (k= 1,2, ..., n−1)

asj→+∞. Then every solutionf 6≡0 of the equation

f(n)+Bn−1(z)f(n−1)+...+B1(z)f0+B0(z)f = 0

has an infinite order.

It is natural to ask what can be said about similar situations in the unit disc ∆. Forn>2,we consider a linear differential equation of the form

f(n)+An−1(z)f(n−1)+...+A1(z)f

0

+A0(z)f = 0 (1.2)

whereA0(z), ..., An−1(z), A0(z)6≡0 are analytic functions in the unit disc ∆ = {z ∈ C : |z| < 1}. It is well-known that all solutions of the equation (1.2) are analytic functions in ∆ and that there are exactlynlinearly independent solutions of (1.2), (see [19]).

A question arises : What conditions onA0(z), ..., An−1(z) will guarantee that

every solution f 6≡0 of (1.2) has infinite order? Many authors have investigated the growth of the solutions of complex linear differential equations in C, see [2, 3,

4, 5, 6, 10, 15, 21, 22, 28, 29]. For the unit disc, there already exist many results [8, 9, 11, 12, 19, 20, 23], but the study is more difficult than that in the complex plane, because the efficient tool of Wiman-Valiron theory doesn’t work in the unit disc.

In the present paper, we investigate the growth of solutions of the equation (1.2). We also study the growth of solutions of non-homogeneous linear differential equa-tions. We give answer to the question posed above and we obtain four theorems which are analogous to those obtained by the author in the complex plane (see [2, 5]).

Theorem 1.7. LetA0(z)6≡0, ..., An−1(z)be analytic functions in the unit disc∆.

Suppose that there exist a sequence of complex numbers(zj)jNwith lim

(4)

and three real numbersα, β andµ satisfying06β < α andµ >0 such that

|A0(zj)|>expp

α

1

1− |zj|

µ

(1.3)

and

|Ak(zj)|6expp

β

1

1− |zj|

µ

(k= 1, ..., n−1) (1.4)

asj→+∞,wherep>1is an integer. Then every solution f 6≡0 of the equation (1.2) satisfies ρp(f) =ρM,p(f) = +∞andρp+1(f) =ρM,p+1(f)>µ.

Settingp= 1 in Theorem 1.7, we deduce the following result.

Corollary 1.8. Let A0(z)6≡0, ..., An−1(z) be analytic functions in the unit disc

∆.Suppose that there exist a sequence of complex numbers(zj)jNwith lim j→+∞|zj|= 1− and three real numbersα, β andµsatisfying 06β < αandµ >0 such that

|A0(zj)|>exp

α

1

1− |zj|

µ

and

|Ak(zj)|6exp

β

1 1− |zj|

µ

(k= 1, ..., n−1)

as j →+∞. Then every solution f 6≡ 0 of the equation (1.2) satisfies ρ1(f) = ρM,1(f) = +∞ andρ2(f) =ρM,2(f)>µ.

Theorem 1.9. LetA0(z)6≡0, ..., An−1(z)be analytic functions in the unit disc∆

withi(A0) =p(16p <+∞)andmax{ρM,p(Ak) :k= 1, ..., n−1}6ρM,p(A0) = ρ.Suppose that there exist a sequence of complex numbers (zj)jNwith lim

j→+∞|zj|= 1− and three real numbersα, βandµsatisfying06β < αandµ >0 such that for any given ε >0 sufficiently small, we have

|A0(zj)|>expp

(

α

1

1− |zj|

ρ−ε)

(1.5)

and

|Ak(zj)|6expp

(

β

1

1− |zj|

ρ−ε)

(k= 1, ..., n−1) (1.6)

as j →+∞. Then every solution f 6≡ 0 of the equation (1.2) satisfies ρp(f) =

ρM,p(f) = +∞andρp+1(f) =ρM,p+1(f) =ρM,p(A0) =ρ.

From Theorem 1.9, by settingp= 1 we obtain the following result.

Corollary 1.10. Let A0(z)6≡0, ..., An−1(z)be analytic functions in the unit disc

∆ with max{ρM,1(Ak) : k = 1, ..., n−1} 6 ρM,1(A0) = ρ < +∞. Suppose that

there exist a sequence of complex numbers (zj)j∈N with jlim+|zj|= 1− and three

real numbers α, β andµ satisfying 0 6β < α and µ >0 such that for any given ε >0 sufficiently small, we have

|A0(zj)|>exp

(

α

1 1− |zj|

(5)

and

|Ak(zj)|6exp

(

β

1

1− |zj|

ρ−ε)

(k= 1, ..., n−1)

as j →+∞. Then every solution f 6≡ 0 of the equation (1.2) satisfies ρ1(f) = ρM,1(f) = +∞ andρ2(f) =ρM,2(f) =ρM,1(A0) =ρ.

In the next, we study the growth of non-homogeneous linear differential equation

f(n)+An−1(z)f(n−1)+...+A1(z)f

0

+A0(z)f =F(z) (n>2), (1.7)

whereA0(z), ..., An−1(z), F(z) are analytic functions in the unit disc ∆ ={z∈

C:|z|<1}. We obtain the following results:

Theorem 1.11. Let A0(z), ..., An−1(z) and F(z) be analytic functions in the

unit disc ∆ such that for some integer s, 1 6 s 6 n−1, satisfying i(As) = p (16p <∞) andmax{ρp(Aj) (j=6 s), ρp(F)}< ρp(As). Then every admissible

solution f of (1.7) withρp(f)<+∞ satisfiesρp(f)>ρp(As).

Settingp= 1 in Theorem 1.11, we deduce the following result.

Corollary 1.12. Let A0(z), ..., An−1(z) and F(z) be analytic functions in the

unit disc ∆ such that for some integer s, 1 6 s 6 k−1 satisfying max{ρ1(Aj) (j6=s), ρ1(F)} < ρ1(As)<+∞. Then every admissible solutionf of (1.7) with

ρ1(f)<+∞satisfiesρ1(f)>ρ1(As).

Theorem 1.13. LetA0(z), ..., An−1(z)andF(z)be analytic functions in the unit

disc ∆ such that for some integer s, 0 6s 6k−1, we have ρp(As) = +∞ and max{ρp(Aj) (j6=s), ρp(F)}<+∞ (16p <∞). Then every solution f of (1.7)

satisfiesρp(f) = +∞.

From Theorem 1.13, by settingp= 1 we obtain the following result.

Corollary 1.14. Let A0(z), ..., An−1(z) and F(z) be analytic functions in the

unit disc ∆ such that for some integer s, 0 6s 6k−1, we have ρ1(As) = +∞

andmax{ρ1(Aj) (j 6=s), ρ1(F)}<+∞. Then every solution f of (1.7) satisfies ρ1(f) = +∞.

Remark 1.15. A similar results to Theorem 1.11 and Theorem 1.13 in the plane case were obtained by the author in[5].

2. Some Lemmas

In this section we give some lemmas which are used in the proofs of our theo-rems.

Lemma 2.1. ([13], Theorem 3.1) Letkandj be integers satisfyingk > j >0, and letε >0andd∈(0,1). Iff is a meromorphic in∆ such thatf(j) does not vanish identically, then

f(k)(z) f(j)(z)

6

1

1− |z|

2+ε

max

log 1

1− |z|, T(s(|z|), f)

!k−j

(|z|∈/E1),

(6)

where E1 ⊂ [0,1) is a set with finite logarithmic measure

R

E1 dr

1−r < +∞ and

s(|z|) = 1−d(1− |z|).Moreover, ifρ1(f)<+∞, then

f(k)(z) f(j)(z)

6

1

1− |z|

(k−j)(ρ1(f)+2+ε)

(|z|∈/ E1), (2.2)

while if ρp(f)<+∞for somep>2, then

f(k)(z) f(j)(z)

6

expp1

(

1 1− |z|

ρp(f)+ε)

(|z|∈/E1). (2.3)

Lemma 2.2. ([19]) Letf be a meromorphic function in the unit disc ∆, and let k>1 be an integer. Then

m

r,f (k)

f

=S(r, f), (2.4)

whereS(r, f) =Olog+T(r, f) + log(11r), possibly outside a setE2⊂[0,1)with

R

E2 dr

1−r <+∞. If f is of finite order (namely, finite iterated 1−order) of growth,

then

m

r,f (k)

f

=O

log( 1 1−r)

. (2.5)

In the next, we give the generalized logarithmic derivative lemma.

Lemma 2.3. Letf be a meromorphic function in the unit disc∆ for whichi(f) =

p>1 andρp(f) =β <+∞, and letk>1 be an integer. Then for anyε >0,

m

r,f (k)

f

=O expp2

1

1−r

β+ε!

(2.6)

holds for allr outside a setE3⊂[0,1) with

R

E3 dr

1−r <+∞.

Proof. First fork= 1.Sinceρp(f) =β <+∞,we have for allr→1−

T(r, f)6expp1

1

1−r

β+ε

. (2.7)

By Lemma 2.2, we have

m r,f f

0!

=O

ln+T(r, f) + ln( 1 1−r)

(2.8)

holds for allroutside a setE3⊂[0,1) with

R

E3 dr

1−r <+∞.Hence, we have

m r,f f

0!

=O expp2

1

1−r

β+ε!

, r /∈E3. (2.9)

Next, we assume that we have

m

r,f (k)

f

=O expp2

1 1−r

β+ε!

, r /∈E3 (2.10)

for some an integerk>1.SinceN r, f(k)

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Tr, f(k)=mr, f(k)+Nr, f(k)

6m

r,f (k)

f

+m(r, f) + (k+ 1)N(r, f)

6m

r,f (k)

f

+ (k+ 1)T(r, f)

=O expp2

1

1−r

β+ε!

+ (k+ 1)T(r, f) =O expp1

1

1−r

β+ε!

.

(2.11) By (2.8) and (2.11), we again obtain

m

r,f (k+1)

f(k)

=O expp2

1

1−r

β+ε!

, r /∈E3 (2.12)

and hence,

m

r,f (k+1)

f

6m

r,f (k+1)

f(k)

+m

r,f (k)

f

=O expp−2

1

1−r

β+ε!

, r /∈E3. (2.13)

Lemma 2.4. ([1]) Let g : (0,1)→R andh: (0,1)→R be monotone increasing functions such that g(r) 6 h(r) holds outside of an exceptional set E4 ⊂ [0,1)

of finite logarithmic measure. Then there exists a d ∈ (0,1) such that if s(r) = 1−d(1−r), theng(r)6h(s(r))for allr∈[0,1).

Lemma 2.5. ([20]) LetA0(z), ..., An−1(z) be analytic functions in the unit disc

∆. Then every solutionf 6≡0 of (1.2) satisfies

ρM,p+1(f)6max{ρM,p(Ak) :k= 0,1, ..., n−1}. (2.14)

Proof of Theorem 1.7. Suppose thatf 6≡0 is a solution of (1.2) withρp(f)<+∞. Then by Lemma 2.1, there exists a setE1⊂[0,1) with finite logarithmic measure

R

E1 dr

1−r <+∞such that for allzsatisfying|z|∈/E1and fork= 1,2, ..., n, we have

f(k)(z)

f(z)

6 1 1− |z|

k(ρ1(f)+2+ε)

(|z|∈/E1), (2.15)

ifρ1(f)<+∞,while ifρp(f)<+∞for somep>2, then

f(k)(z)

f(z)

6

expp1

(

1

1− |z|

ρp(f)+ε)

(|z|∈/ E1). (2.16)

By (1.2), we can write

16 1

|A0(z)|

f(n) f +

An−1(z) A0(z)

f(n−1) f +...+

A1(z) A0(z)

(8)

Then from (1.3), (1.4), (2.15), (2.16) and (2.17), we next conclude that

16 1

expnα 1 1−|zj|

µo

1

1− |zj|

n(ρ1(f)+2+ε)

+ exp

(β−α)

1

1− |zj|

µ 1

1− |zj|

(n−1)(ρ1(f)+2+ε)

+...+ exp

(β−α)

1

1− |zj|

µ 1

1− |zj|

ρ1(f)+2+ε

6n

1

1− |zj|

n(ρ1(f)+2+ε)

exp

(β−α)

1

1− |zj|

µ

(2.18)

or

16 1

expp

n

α1−|1z j|

µoexpp−1 (

1 1− |zj|

ρp(f)+ε)

+

expp1−|1z j|

µo

expp1−|1z j|

µoexpp−1 (

1 1− |zj|

ρp(f)+ε)

+...+

expp1−|1z j|

µo

expp1−|1z j|

µoexpp−1 (

1 1− |zj|

ρp(f)+ε)

6n

expp1−|1z j|

µo

expp1−|1z j|

µoexpp−1

(

1 1− |zj|

ρp(f)+ε)

(2.19)

holds all zj satisfying|zj|∈/ E1 as |zj| →1−, both the limits of the right hand of inequalities (2.18) and (2.19) are zero as |zj| →1−. Thus, we get a contradiction. Henceρp(f) =ρM,p(f) = +∞.

Now letf 6≡0 be a solution of (1.2) withρp(f) = +∞.By Lemma 2.1, there exist s(|z|) = 1−d(1− |z|) and a set E2 ⊂[0,1) with finite logarithmic measure

R

E2 dr

1−r <+∞such that forr=|z|∈/E2, we have

f(k)(z)

f(z)

6 1 1− |z|

2+ε

max

log 1

1− |z|, T(s(|z|), f)

!k

(k= 1, ..., n).

(2.20) By (1.2), we can write

|A0(z)|6

f(n) f

+|An−1(z)|

f(n−1) f

+...+|A0(z)|

f0 f . (2.21)

Again from the conditions of Theorem 1.7, for all zj satisfying |zj| =rj ∈/ E2 as |zj| →1−, we get from (1.3), (1.4), (2.20) and (2.21) that

expp

α

1

1− |zj|

µ

6|A0(zj)|6nexpp

β

1

1− |zj|

(9)

×

1

1− |zj|

2+ε

max

log 1 1− |zj|

, T(s(|zj|), f)

!n

. (2.22)

Noting thatα > β>0, it follows from (2.22) that

exp

α(1−γ) expp−1

1 1− |zj|

µ

6n

1

1− |zj|

n(2+ε)

Tn(s(|zj|), f) (2.23)

holds for allzj satisfying|zj|=rj∈/E2 as|zj| →1−,whereγ(0< γ <1) is a real number. Hence by Lemma 2.4 and (2.23), we obtain

ρp+1(f) =ρM,p+1(f) = lim rj→1−

suplog

+

p+1T(rj, f) −log (1−rj) >

µ.

Theorem 1.7 is thus proved.

Proof of Theorem 1.9. Suppose thatf 6≡0 is a solution of equation (1.2). Using the same proof as in Theorem 1.7, we getρp(f) =ρM,p(f) = +∞.

Now we prove that ρp+1(f) = ρM,p(A0) = ρ. By Theorem 1.7, we have ρp+1(f)≥ρ−ε, and since ε >0 is arbitrary, we getρp+1(f)≥ρ. On the other

hand, from Lemma 2.5, we have

ρp+1(f) =ρM,p+1(f)6max{ρM,p(Ak) :k= 0,1, ..., n−1}

=ρM,p(A0) =ρ.

It yieldsρp+1(f) =ρM,p+1(f) =ρM,p(A0) =ρ.

Proof of Theorem 1.11. Let max{ρp(Aj) ( j6=s), ρp(F)} = β < ρp(As) = α. Suppose thatf is an admissible solution of (1.7) withρ=ρp(f)<+∞. It follows from (1.7) that

As(z) =

F f(s) −

f(n)

f(s) +An−1(z) f(n−1)

f(s) +...+As+1(z) f(s+1)

f(s)

+As−1(z) f(s−1)

f(s) +...+A1(z) f0

f(s)+A0(z) f f(s)

= F

f(s) − f f(s)

f(n)

f +An−1(z) f(n−1)

f +...+As+1(z) f(s+1)

f

+As−1(z) f(s−1)

f +...+A1(z) f0

f +A0(z)

. (2.24)

By Lemma 2.3 and (2.24), we obtain

T(r, As) =m(r, As)6m(r, F) +m

r, 1 f(s)

+m

r, f f(s)

+ n−1

X

j=0 j6=s

m(r, Aj) +O expp−2

1

1−r

(10)

6T(r, F) +T

r, 1 f(s)

+T

r, f f(s)

+ n−1

X

j=0 j6=s

T(r, Aj)

+O expp−2

1 1−r

ρ+ε!

(2.25)

holds for allroutside a setE3⊂[0,1) with

R

E3 dr

1−r <+∞.Noting that

T

r, f f(s)

6T(r, f) +T

r, 1 f(s)

=T(r, f) +Tr, f(s)+O(1)

6(s+ 2)T(r, f) +o(T(r, f)) +O(1). (2.26) Thus, by (2.25), (2.26), we have

T(r, As)6T(r, F) +cT(r, f) + n−1

X

j=0 j6=s

T(r, Aj)

+O expp2

1 1−r

ρ+ε!

(r /∈E3), (2.27)

where c >0 is a constant. Sinceρp(As) =α, then there exists

n

r0no rn0 →1− such that

lim r0n→1−

log+p Tr0n, As

−log (1−r0n)

=α. (2.28)

Set RE

3 dr

1−r := logγ <+∞. Since

R1−

1−r0n γ+1 r0n

dr

1−r = log (γ+ 1), then there exists a pointrn∈

r0n,1−1−r

0

n γ+1

−E3⊂[0,1). From

log+p T(rn, As)

log11r

n >

log+p Tr0n, As

logγ+1

1−r0n

=

log+p Tr0n, As

log11

−r0n+ log (γ+ 1)

, (2.29)

it follows that

lim rn→1−

log+p T(rn, As)

log 1 1−rn

=α. (2.30)

So for any givenε(0<2ε < α−β),and forj 6=s

T(rn, As)>expp−1

1

1−rn

α−ε

(2.31)

T(rn, Aj)6expp−1

1

1−rn

β+ε

, T(rn, F)6expp−1

1

1−rn

β+ε

(2.32)

hold forrn→1−.By (2.27), (2.31) and (2.32) we obtain forrn→1−

expp1

1

1−rn

α−ε

6nexpp1

1

1−rn

β+ε

(11)

+O expp2

1

1−rn

ρ+ε!

. (2.33)

Therefore,

lim rn→1−

sup log

+

p T(rn, f) −log (1−rn) >

α−ε (2.34)

and since ε > 0 is arbitrary, we get ρp(f)> ρp(As) =α. This proves Theorem

1.11.

Proof of Theorem 1.13. Setting max{ρp(Aj) (j =6 s), ρp(F)}=β,then for a given

ε >0, we have

T(r, Aj)6expp−1

1

1−r

β+ε

(j6=s), T(r, F)6expp1

1

1−r

β+ε

(2.35)

forr→1−.Now by (2.24), we have

T(r, As) =m(r, As)6m(r, F) +m

r, 1 f(s)

+m

r, f f(s)

+ n−1

X

j=0 j6=s

m(r, Aj) + n

X

j=1 j6=s

m

r,f (j)

f

6T(r, F) +T

r, 1 f(s)

+T

r, f f(s)

+ n−1

X

j=0 j6=s

T(r, Aj) + n

X

j=1 j6=s

m

r,f (j)

f

. (2.36)

Hence by (2.26) and (2.36) we obtain that

T(r, As)6T(r, F) +cT(r, f) + n−1

X

j=0 j6=s

T(r, Aj) + n

X

j=1 j6=s

m

r,f (j)

f

, (2.37)

wherec >0 is a constant. Ifρ=ρp(f)<+∞, then by Lemma 2.3

m

r,f (j)

f

=O expp2

1

1−r

ρ+ε!

(j= 1, ..., n;j 6=s) (2.38)

holds for allroutside a setE3⊂[0,1) with

R

E3 dr

1−r <+∞.Forr→1

,we have

T(r, f)6expp−1

1

1−r

ρ+ε

. (2.39)

Thus, by (2.35), (2.37), (2.38) and (2.39), we get

T(r, As)6nexpp−1

1

1−r

β+ε

+cexpp1

1

1−r

ρ+ε

+O expp2

1 1−r

ρ+ε!

(2.40)

forr /∈E3 andr→1−. By Lemma 2.4, we have for anyd∈(0,1)

T(r, As)6nexpp−1

1

d(1−r)

β+ε

+cexpp1

1

d(1−r)

(12)

+O expp2

1

d(1−r)

ρ+ε!

(2.41)

forr→1−.Therefore,

ρp(As)6max{β+ε, ρ+ε}<+∞. (2.42)

This contradicts the fact thatρp(As) = +∞. Hence ρp(f) = +∞.

Acknowledgments. The author would like to thank the referee for his/her helpful remarks and suggestions to improve this article.

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(13)

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Benharrat Bela¨ıdi

Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem (UMAB), B. P. 227 Mostaganem-(Algeria)

References

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