Journal of Mathematical Sciences, Vol. 85, No. 5, 1997

THE G R O W T H OF ONE CLASS OF ENTIRE FUNCTIONS

M . N . S h e r e m e t a

Let A be the class of positive sequences A = (An), and let A(R) be the class of functions f(z) = 7]~n=o~176 anz" that are analytic in the disc {z: Izi < R}, 0 < R < +oo. We say that f E A~(1) i f f E A(1) and tanl <_ laxlA, for all n E N . We assume that (np) is an increasing sequence of natural numbers, and no = 0. It was proved in [1] that for each of the sequences A E A and f E A(1), the condition f("p) E Ax(1) for all p E Z+ implies that f E A(oo) if np+l - n p = O(1) (p --+ ~ ) , and only if lirnp_+oo (np+ 1 - n p ) < +c~. The condition np+l - n p = O(1) (p --+ oo) can be weakened if we impose an additional requirement on the sequence A. Namely [1], for each sequence A 6 A such that lnA, = O(n), n --4 ~ , and each function S E A(cx~), the condition f("D E Ax(1) for al lp E Z+ implies that f e A(oo) if np+i/np -~ 1 (p --+ c~), and only if lir%~oo np+llnp = 1. Assuming that this or some other sufficient condition is satisfied, we will investigate the growth of the corresponding entire function.

T h e o r e m 1. Let A E A, and np+l - n p = O(1) (p -+ oo). We set l = max{np+l - np: p E Z+} and L = max{An: 2 < n < l + 1}. I f f ( " , ) E Ax(1) for Mlp E Z+, then f E A(oo) and

cr = li---m In M(r, f ) < ~ {L(l + 1)!} p/"p, (1) p-@oo r p--+oo

where M(r, f ) = max{l/(z)l: Izl-- r}.

P roo f . It was shown in [1] that if A E A and f("p) E Ax(1) for all p E Z+, then for all p E Z+ and m, 1 <_ m <__ np+l - n p + 1, we have the inequality

' la.,+.. I < lax l (.m!A.,..p P P

_ +m)!l-i(n~-nj_l+l)!l'XA.j_nj_l+l (2) j = l j : l

Since nj - n j_ l + 1 < 1 + 1 and A,i_,i_~+l < L for all j E N , it follows from (2) that we have the inequality

la,,+.~l < la~l{L(l + 1)!}P+i/(np + m)!

We can now use the Hadama.rd formula to obtain

o" = l'im np + m n,,+mllan m I plum p-+oo e V t p+ ' <- {L(t + 1)!}"~+~

where 1 < m < 1 + 1, i.e., we are led to inequality (1). We will show that estimate (1) cannot, in general, be improved. Assume that np = pl (p E Z+, l E N) , An --

L = (l + 1 ) l / ( / + 1)!, n > 2, and consider the function

x-~ ( l + 1 ) pl . ~_, f l (z) C.;:cg, z ' -

For this function the Hadamard formula yields

cr = li---~ (l + 1) 'p/"~ = ~ {L(l + 1)!} p/"p, p--+oo p--4oo

and it remains to show that f(n~) E Ax(1) for all p E Z+. Since

(l + 1)(p+k) l . . . . . . Znp+k --np+l f}nP)(z) = E (np+,~ -np + 1)!

k=O

we must show that (l + 1) kL < n(np+k - n p + 1)! = (l + 1)t(kl + 1)!/(l + 1)!

for all k E N . The last inequality is obvious if k = 1, and it can easily be proved for all other k E N by induction. It follows from Theorem 1 that if np+~ - n p = O(1) (p --+ c~), then for each sequence A E A, the condition

f(n,) E A~(1) for all p E Z+ implies that the order of the function f is no greater than 1. If, however, np+l - - np --+ cx~ (p ~ co), f ( " , ) E Ax(1) for ~11 p E Z+ and f E A ( ~ ) , the order may be greater than one. However, we can also obtain an estimate of the order in this ease.

Translated from Teoriya Funktsii, Funktsional'nyi Analiz, i [kh Prilozheniya, Vol. 58, pp. 121-125, 1993. Original article submitted August 8, 1991.

1072-3374/97/8505-2235518.00 �9 Plenum Publishing Corporation 2235

T h e o r e m 2. Let A 6 A and lnAn = O(n) (n -+ oo), and let np+l/np --+ 1 (p -+ oo). l f f (",) 6 Ax(1) for allp 6 Z+. then f E A(oo) and

p = li--'m l n l n M ( r , f ) < q = 1 ~ lnnp ~-+oo In r - p~oo In 1 (3)

n ~ , / n p _ , - 1

Proof . In the proof of Theorem 4 of [1] it is shown that ifA 6 A, lnAn = O(n) (n --+ oo) and f("~) 6 Ax(1) for all p 6 Z+, then for all p E N and m, 1 <_ m <_ np+l - n v + 1, we have the inequality

c~+1 } 1 in ____~1 C--2-P, In np+l - - - + 0 ( 1 ) (4) np +-------m [a,,~+m[ >- min lnnv - np np+l

when p --+ c~, where P

Cp = E ( n . i - n j -1) ln(nj - nj-1). j = l

Since nv+l/np --+ 1 (p --+ oc), we have lnnp .., ln(n 0 + m) ~ lnnv+l (p --+ or for any m, 1 < m < nv+l - np + 1. As a result, the Hadamard formula for finding the order p of a function f yields

1 1 1 - = lim I n - p p~oo (np+ m) In(rip + m) [a.,+m[

( > ~imoo 1 npl--~np = 1-pli~m~n,-~nn, " (5)

If, now, q = +oo, inequality (3) is obvious. However, if q < oo, then, for any e > 0 with p > P0 (e) we have

l n . , <_ (q + ~ ) l n { 1 / ( n ; ~ _ ~ - ~}, i.e., in( . , - . , _ ~ ) _< (1 - ~ ) l . . , . as a %

result

( ) p Cp < 1 I InnpE(n j - n j _ l ) + O ( 1 ) -- q 7 E j = l

(1 1 ) = np lnnp+O(1) (V-+Co). q + e

Inequality (3) now follows from (4), because ~ is arbitrary. We will now show that in general, estimate (3) cannot be improved. Since q > 1 for any increasing sequence (n0)

of natural numbers, and for the case q = 1, the corresponding example of a function f l of order one has alreMy been constructed, we assume that q > 1.

r 1-1/q~ Let n 0 = 0 , n l = l , a n d n p = n p - l + t n v _ l l ,P>-2 , and

S2(z) = Z ~ z""§ p=0

{1 } a p = e x p - q ( n p + l ) l n ( n p + l ) .

By Hadamard's theorem, the function f2 has order p = q, and it remains to show that f~'*~) E Ax(1) for all p 6 Z+ and some sequence A 6 A such that In An = O(n) (n ~ oc). Since

o o

f~'~')(z) = E (n_..~p+~ + 1)! ~:1 (n~+~ - ~ u 1)! %+~z~+'-""+1'

we must show that

i.e.,

(nv+k + 1)! ap+k < X,~+k_,,+l '

(nv+k -- n v + 1)!(n v + 1)! a v --

ln(nv+k + 1)!+ lnav+k - ln(n v + 1 ) ! - lna v

- ln(nv+k - n v + 1)! _< M(np+k - n v + 1)

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for all p E Z+ and k E N , where M is some positive constant. Using Stirling's formula, we can show without difficulty that the last inequality is satisfied if

(1- ~ ) {(nv+k + 1) ln(nv+k + 1 ) - (np + 1) ln(np + 1)}

-(nv+k - n v + 1) ln(nv+k - n p + 1) - M(nv+u - np + 1) < 0 (6)

for all p E Z+ and k E N . For p = 0 and k E N inequality (6) is obvious. As a result, we must prove it for p E N and k E N .

On [nv+l + 1, oo), consider the function

r

we will show that

r = ( 1 - ~ ) l n z - ln(z - nv) - M - 1 / q < O

for appropriate M > 0. Indeed, the function r is a decreasing function on [nv+z + 1, oc) and

r + l) = ( 1 - ~ ) ln(nv+z + l) - ln([@-l/q] + l)

- M - 1 < ( 1 - ~ ) l n n v + l + ( 1 - ~ ) l n 2 - - M q np

- - l < 2 ( l - ~ ) l n 2 - M - l - < 0 q - q

if M is sufficiently large. As a result, the function r is a decreasing function on [np+l + 1, oo), and it is sufficient to prove inequality (6) for

k = l and a l l p E N . But (r~p+ 1 + 1 ) l n ( n p + l + 1) -- (rip --b 1)ln(nv + 1)

and

< (np+l - nv)(ln nv+l + 2) < (rip+ 1 - - np)(ln nv + 2 + In 2)

(np+l--np-b l)ln(np+l--np + l) >_ (rip+l--rip) (1- -~) lnnp.

As a result, inequality (6) is satisfied for k = 1 and p E N if (1 - 1/q)(2 + ln2) < M.

Thus, if Ati = exp{(2 + ln2)n}, then f~tiP) E Ax(1) for all p E Z+, and the sharpness of (3) is proved.

R e m a r k 1. If all the conditions of Theorem 2 are satisfied and we also have lirr~_.~ ~P - 1 < +eo, then t ip- - I

= ~o < + o o . Indeed, i n t h i s c a s e n p - n v _ l < a n p _ 1 ( p E Z + ) , O < d , eo, Cp<_ 1 - n v l n n p +

np In d, In np - ~ > - 1 In np - In d, and it follows from(4), by Hadamard's formula, that we have t i p - -

lira np + m __.e.._ = - - [anv+ml ""+" < K lira nv+l < oo, p~oo ep p-+oo n v

where K is some positive constant. R e m a r k 2. Theorem 2 can be extended to the case of generalized order p~z(f) = li--m,.-~oo c~(ln M(r, f ) ) / f l ( ln r). Let L ~ be the class of all functions o~ that positively increase to +0% are continuously differentiable on [a, +oo),

and are such that a(x(1 + o(1))) ~ a(x) (x -+ +oc). We say that a E A ~ if a E L ~ and a(cx) ~ a(x) (X --+ +oc) for every c, 0 < c < +oo. It is known [2] that if a E A ~ fl E L ~ and

d ~ - l ( c ~ ( x ) ) - O(1) (x ~ + o o ) , 0 < c < + o ~ , (7) d lnx

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then

p~a(f)---,~li---m o~(n)/~ ( 1 In i-~1) . (8)

We now write qa# = ti----m,_+oo o~(np)/Z(ln{1/(np/np_l)}). Then it is not difficult to use (4) and (9) to prove the following assertion: If a E A ~ fl E L ~ condition (7) is satisfied, and the conditions of Theorem 2 are satisfied, then

pe,#(f) < qaz.

References

1. A .A . GOL'DBERG and M. N. SHEREMETA, "On analytic continuation of functions that are analytic in a disk to the entire plane," Teoriya Funktsii, Funktsional'nyi Analiz, i Ikh Prilozheniya, 58, 21-30 (1993).

2. M.N. SHEREMETA, "The connection between the growth of the maximum modulus of an entire function and the moduli of the coefficients of its power expansion," Izv. Vuzov, Set. Mat., No. 2, 100-108 (1967).

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