Internat. J. Math. & Math. Sci.
VOL. 12 NO.
(1989) 15-28
15
GROWTH OF ENTIRE FUNCTIONS WITH SOME UNIVALENT
GELFOND-LEONTEV DERIVATIVES
G.P. KAPOOR, O.P. JUNEJA and J. PATEL
Department of Mathematics
Indian Institute of Technology, Kanpur
Kanpur, 208016, India
(Received December 2, 1985 and in revised form July 21, 1986)
I.
INTRODUCTION.
In= 0 anZ
f(z)
Let
{d
sequence of positive numbers
n
be analytic in
z
<
R.
For a non-decreasing
the Gelfond-Leontev (G-L) derivative of f is
n n=l
defined as [I]
Z
Df(z)
Dkf,
The kth iterate
k=1,2,..., of D is given by
Dkf(z)
En=kdn...dn_k+lanz
e
En= k
n-k
e
f(z)
0
anZ
n
Lf(z)
into
n-k
(1.2)
n-k
n
e--(dld
r.n=
a z
n
e =I and
...d
2
o
n
n
derivative of f; whereas, if d
n
where,
(I.I)
d a z
n n
n=l
-I
=-
If
d =-n, Df is the ordinary
n-l,2,
n
I, D is the shift operator L which transforms
r..n=lanz
n-I
Let,
(z)
and have radius of convergence
R
o
(1.3)
ln=oe n
lim
n
d
n
R
o
From the monotonicity of
{d
n n=l
we have
sup {d }.
n
nl
(z).
Thus, (z) bears the same relationship to the
operator D that the function exp(z) bears to the ordinary differentiation.
For an entire function f, Nachbin used the function (z) as a comparison
r.
Thus, the
function for measuring the growth of maximum modulus of f on
Clearly,
(0)
and
D(z)
Izl
G.P. KAPOOR, O, P. JUNEJA AND J. PATEL
16
-type of f is defined as the Inflmum of the positive numbers
growth parameter
such that, for sufflclently large
If(z)
where,
z(f).
r,
M(r)
(1.4)
(z) is entire and M is a positive constant.
It is known [2,p.6] that
-type of f as
We denote
a
n
n
/
d
n, the -type of an entire function f reduces to its classical exponential
n
type and the formula (1.5) gives its well known coefficient characterlsatlon [3, p.
For
11].
(z) can also be used to define a measure of growth
The comparison function
analogous to classlcal order [3, p.8] of an entire function.
Thus, for an entire
-order p(f) of f be defined as the inflmu of positive numbers
p such that, for sufflclently large r,
function f, let the
K(r p)
If(z)l
(z)
where
is entire and K is a positive constant.
[4,5] showed that
Shah and Trimble
classical
(1.6)
f(np
derivatives
(n
increasing sequence
are
if f is entire then, the assumption that the
unlvalent
A- {z:
in
[z[ <
I} for
a
sultable
of positive integers affects the growth of the maximum
p
n
instead, we asse that the G-L derivatives D Pf of an entire
function f are unlvalent in
A, then it is natural to enquire in what way the
modulus of f.
-type and
If
-order of f are influenced.
The present paper is an attempt In this
n
In Theorem 1, we find that if f is entire,
sup (rip-rip_ I)
< ", then the -type
direction.
lira
p/
,
-=,
n
if f is entire,
D
(f)
then f need not be of finite
Pf
are univalent in
P(f)
lim
sup
p/=
It is clear that if
n
between
this
D
Pf
nature
0
and the
exists.
Pf
are univalent in
and
of f must satisfy
2(d(l+1).,.d(2))
’(f)
Further, if
D
p(f)
A and
n
p
-type.
n
p+l
is
as
p
,
then
ioE d(np-np_ I)
log d(n
p
1, then the above inequality gives no relationship
-order of an entire function f.
This
Our Theorem 2 shows that
illustrated
in
Theorem
In fact, no such relation of
3, wherein for any given
GROWTH OF ENTIRE FUNCTIONS ITH UNIVALENT DERIVATIVES
p, 0
4
I, and any given increasing sequence
P
[np}p.
of positive integers, we
n
O-order p, such that
construct an entire function h, of
and only if
-TYPE AND
d
n
Let
En= 0 a n z
f(z)
n
sequence
Suppose
of positive integers.
,
(np-np_ I)
sup
llm
P
satisfies
2(dC+l)...d(2))
T#Cf)
Le___t
<
D
Pf
".
Pf(z)
}.I
be an
be analytic and univalent in
Then_____, th___e
-type
T (f)
o__f
f
(2.1)
Zk= 0 d(n?k)...d(k+l)a(np+k)z
a.
are univalent in
Since, for any function
d
la(np+k)
4 k
d
d
p, we get, for
p
2
2 and
)
d
la(np+k)]
Ak
k)
k..
.dl
U’, U’
and inducting upon
i=2 (ni-ni-l+l)d(ni-ni-l+l)’’’d(2)
(2.3)
Hence, for sufficiently large p,
l/(np+k)
(I+o(I))(dk...d
(d
k=np+l-n/l
p
.d
A=d(nl+l)...d(2)la(nl+l) I.
a
(2.2)
np+l-np+l
k
dk+n" ..1.dl
k.
univalent in
P
P
[e(n(np+P+k)]
d(np+l)...d(2)la(np+l)
In particular, putting
for k=l,2,.., and p=2,3,
bo+blZ+b2z+...
get
k. .d
k+n
k
G(z)
Ibnl ’
A ,it is known [6] that
(np+l-np)
p
1/
n
D
Since,
{n
By the hypothesis,
PROOF.
where
.
as n
be an entire function and
n
A
is univalent in
EXPONENTS OF UNIVALENT G-L DERIVATIVES.
THEOREM I.
Increasing
Ph
n=n
P
In the sequel, we shall assume throughout that
2.
D
>
1/(n +k)
P
i=2
1/(np+k)
1/(np+k)
p
{(nl-ni_1+l) d(ni-ni_l+l)
d(2)
is an increasing function of k, and
U, for sufficiently large p,
I/(n +k)
(dk...dl)
P
(d(np+l_np+l)...d(1))
I/np+ I=
(I+(I))
18
G.P. KAPOOR, O.P. JUNEJA AND J.
Further
[7], for
2
p
P
II
PATEL
I/(np+2)
(ni-ni_l+l)
p/n
n
(I+
p-)
P 2
inequallty in (2.4), we get for sufflclently large p,
Using (2.5) and the preceding
-
1/(n +k)
P
a(n_+k)
p
2(I+I))
Now, if
O,
aj
tj
0, r.
tj >
l[
(d(nl-ni_l +I)...d(2)
0 and
max
IJN-I
()
N
Further, log(d(J+l)...d(2))/J is an increasing function of
j
Thus, if
1,2,
J
,
,
Suppose
P
for
nj+l-n j
J
Ji"
J
I/(rip+k)
(2.6)
then clearly,
(2.7)
for
.log(d(+l)...d(2))
log(d(J+l)...d(2))
Let
(2.5)
tT
(2.8)
is the number of
Ji’s
in
[po,p]
such that
Then, by (2.7)and (2.8),
P
.z .d(2))
Po11og(d(nj_nj_l+l).
r.
Po+l
t
7___’1
p
(nj-nj_
7
r. y
7=I
(log(d(7+l)
d(2)
log(d(+I.)....d(j2)
U
t7
The above inequality implles that
p
II
I=2
(d(nl-n i-I+
I/(np+k)
l)...d(2))
P
exp
I_Z2 log(d(ni-nil+l)...d(2)
(2.9)
np
P
exp {o(I)+
Pol log(d(n -ni-I +l)...d(2))-}
i
p
Po +Z1 (nl-ni-
!og(d.(.U+1)...d(2).)}
exp(o(1)+
Using the estimate (2.9) in (2.6) and proceeding to llmlts
a
lim
k/(R)
sup
I/k
k
J-k
,a(n +k)
1/(rip+k)
{le (np+k)l
p
2(d(+l)...d(2)) 1/.
llm sup
This completes the proof of the theorem.
2
k
np+ l-np+l,
p
2)
GROWTH OF ENTIRE FUNCTIONS WITH UNIVALENT DERIVATIVES
In Theorem I, it is sufficle.nt to take the function f to be analytic in
in the definition of G-L
if the sequence [d
R, for some R, 0 < R <
n
REMARK I.
Izl
<
19
,
n=
derivative of f satisfies the condition
m
((r-t_ 2
ltm
logd(1) )/m) -.
In fact
for an
n
,
sup (n -n
p p-I
<
m
El= 2
lira
", and
log d(1)
To see this, we use (2.5) and
holds, then f is necessarily entire.
(d k. ..d I)
1/(n +k)
p
I+o(I)
for sufficiently large p in (2.3) to get
a(np+k)
1/(np+k)
(2.10)
n
)exp [1_._
n
2( 1+o(
p
p
i2
lg(d(ni-ni-I +l)...d(2))
n +k
P
E
log d(i)]
i-2
p
for sufficiently large p. But since, for sufficiently large
n +k
p,(np-np_
n
0 as p
iZ2_ log(d(ai-ni_l+l)...d(2))
P
Thus, by (2.10) and the condition
ltm
sup
k/
lakl
llm
I/k
lira sup
((r.= 2
1ogd(i))/m)
{[a(np+k)l,, I/(n/k)
2
k
’ np+l-n p+1
P
2}
0.
REMARK 2.
The inequality (2.1) can be improved by imposing suitable additional
{d
restrictions on the sequence
n
}n-i"
For example, let the sequence
{dn}:
be such that
{d(n+2)}n
3 n+l),
d(n+l)...d(2)
Note that (2.11) is satisfied for
d
n
n-1,2,3
=na,a
I.
Because of (2.11), the function s(J) defined by
s(j)
log(d(J+l)...d(2))+log(J+1)
is an increasing function of
J
J
and so for
j=1,2,...; V=I,2...
(2.11)
G.P. KAPOOR, O.P. JUNEJA AND J. PATEL
20
log(d(+1),, ,d(2))+1og(/1_)_
1og(.d(.J+1),, .d(2).)+log(J+l)
j
Let
ty
be the same as in the proof of Theorem 1.
p
II
{(ni-nl_l+l)d(nl-ni_l+1)...d(2)}
i=2
Using (2.7) and (2.12), we get
1[ (n +k)
P
P
exp {o(I) +
Pol
{log(d(nl-ni_ l+l)...d(2))
The above inequality, when employed in
a
I/(rip+k)
l(n(np+k)
+k)
p
(2.12)
U
+
log(nl-ni_l+1)}
(2.4), gives
p
(I+O(I)) lI
{(nl-nl_l+l)d(ni-nl_l+l)...d(2))
i-2
I/(n +k)
P
(+1)11(d(N+1)...d(2)) I/.
Now, on proceeding to limits, we get
(+1)
x(f)
lip
It is clear that the bound on
REMARK 3.
proved in [8].
By taking
shoes that if
Theorem
in (2.13) is better than that in (2.1).
gives
=I, Theorem
2d(2), a result recently
x(f)
x(f)
(np+l-n p)
+
is of finite
sup(np-np_ I)
,
-type.
then f need not be of
Let {n
be an increasing sequence of positive integers such that
p p=l
2 for all p. Further, assume that the sequence {d }m
is such that
n n=l
=-
and log d(n)
(i)
d
(ll)
n
p
o(n
P
=O(nplog
(ill)
lim
P
-type.
EXAMPLE.
0(I), then f
(np-np_ I)
We now give an example to show that if
finite
(2. t3)
(d(+1)...d(2))
p
d(n ))
P
log n as n
GROWTH OF ENTIRE FUNCTIONS WITH UNIVALENT DERIVATIVES
n
where,
Let
r-=
P
+1)...d(Z) ),
log(d(ni-ni_
2
21
(nl)-(n2)
be a non-decreasing step function such that
exp (n)
(n
P
2
q(x)
(n
and
2
p
p-I
n
p
p
<
n
p+l.
x
Let
(j)
d(J+l)...d(2) (J-n +1)
J-np
if
P
for some p
gJ+l
0
otherwise.
Define
g(z)
Zj=og j z
j
We have
We first show that g is an entire function.
suplgkl I/k=
lira
k
(n
[d(nl
lira sup
p
/
exp(n /n
lira
sup[
p
I/n +I
p
)...d(
P
P
P
(d(np+l)...d(2)) I/np+l
n +I
n
lira sup [exp
p/(R)
Since log
(p
p
n +I
P
log d(1))]
iE-2
using the condition (ill), we get from the above
d(n) ~log n as n
inequality that
imsup
Hence g is
entire,
’lgkl z/"
0.
It is easily seen that g is of order
1,
But, by the condition
(li),
llk
9(n_)
lira sup
lira sup
p+
k+(R)
l im sup
Thus, f is not of finite
n
D
are univalent in
Pg(z)
A
@-type
(e( n,,+, )dlnI;-t"l)-..-. d(2) )l/n P
-I-1
exp(n_Inp)
P
2
It remains to see that
Zk= d(np+k+l)-..d(np+k-np+2)a(np+k+l)z
To this end, it is enough to prove that
d(np+k+l .d 2
1(rip+k-rip+l)
Zk.
d(np+k-n/1)...d(2)-
np+k-np+l
22
G.P. KAPOOR, O.P. JUNEJA AND J. PATEL
d(n
p
/l)...d(2),,la(np/l)l
or, equivalently to show that
(n
p+k
Zk-I d(np+k-np+l).., d(2) $(np).
the last inequality reads as
Using the definition of
exp(p+k-p)
(2.14)
I.
Zk= 2-- d(np+k_np+1)...d(2
Now, an induction on k, gives, for k=1,2,3,...
p+k
(p+k-np) pl d(ni-ni-1 +l)...d(2) d(np+k-np+l)....d(2)
exp
Hence, (2.14) is clearly satisfied.
Y-ORDER AND EXPONENTS OF UNIVALENT G-L DERIVATIVES.
3.
A function S(x), continuous on
for every positive number c > 0,
[1,’),
is said to be Slowly Oscillating
(S.O.) if
S(cx)
S(x)
lim
A function H(n) is said to be the restriction of a Slowly Oscillating function
S(x) if S(n)
H(n) for every positive integer n. It is known [9] that, as k
z ki-l H(1)
THEOREM 2.
{n
Let
k(k).
Z
f(z)
n
a z
be an entire function of
be a strictly increasing sequence of positive integers.
p
A, such
analytic and univalent in
n
p
-
d(np-np_
l-llmp (R)sup [log d(np)
order
O and
n
Let
D
Pf
be
If log d(n) 18 the
as p /
p+l-restriction of a slowly oscillating function on integers, then
that
(3 1)
n
(3.2)
log
-]
/
We need the following lemmas.
LEMMA I.
Let
be defined by (1.3).
Then,
n(l
Yn
PROOF.
n
Since
n-k
n n
eked
e d
n n
{d)
n
--)
en_aa
).
Let
Y n =mln
xY0
(xa)x -n
a
>
0
(3.3)
n--I is increasing, we note that for any pair of integers k and
Thus,
GROWTH OF ENTIRE FUNCTIONS WITH UNIVALENT DERIVATIVES
*(x a)
Let
0
<
ak
ak
x
endnn kO d-k
n
!
<
w
-k 0 e kx
Setting
x
=w nl’a
we get
W
-n
w
--)
(x:)x:n en dn(1
n
(
Choosing
23
w- (n/n+a)
I/a to
mlnlmle the lht-hand ide ol the above inelallt2, we
have
Yn
LE 2.
4
n
0<I
t f(z)
where the sequnce {d(n)}
n(1
,(xC)x;n
En.Df0 a sz
i
(
1
e d
n n
(+a),)
be an entire function of
such that log d(n)
s
-order
the restriction
o slowly
oscillating function oposltlve integers.
n log _O(nl
llm sup" log
lan
p,(f)
PROOF.
(3.4)
By Cauchy’s inequality, we get
Since f is of
-order
p#(f) =-
> O,
p, for any
M#(rP+).
If(z)
So that
Using Lemma
I, we have
n(1
I%1 ’ " endn
+)
e(n+p+)
(-’(’P+)-"
(3.5)
But, since log d(n) is the restriction of a S.O. function, by (3.1),
log d(1)
=.
n log d(n) as n +
.up
n
!0g d(n)
Thus, it follows from (3.5)
’
To prove that equallty holds in (3.4), suppose that
lim sup
n
Then, there exist
for
Izl
P
n.
log d(n)
log
<
p such that
iani
<
P.
lanl
<
I/P
e
n
for
n
>
n
o
It now follows that,
r,
n
(3.6)
(0(I) + n +I e n
O
I/Pt
r
n
4
G.P. KAPOOR, O.P. JUNEJA AND J. PATEL
Choo8 e
lo r
N(r)
It is easily seen that
-n
en < ekdt
+
=.
as r +
Since for all values of k and n,
we have
Zn.0
I/Pl
e
r
n
n (
1/P
Zn-O
k-n/P
d
ek
lip
dk
r
k
lJo
ek
n
(’
d
Let k be chosen
I/01 <
(r/d k
such that
"--o
I/Pl
"
e
k
(d
k
I. Then,
.k+I/p.
a
n
;iO )n"
I/p,
k
(3.7)
r)
k
Since the left hand side of (3.7) is independent of k, letting
1/1
En=0 e n
r
n
<
k
/
", we get
1.
us
En-N(r)
Since,
en1101
rnfo(l),
),
@(r
exp (N(r)log r)
r
as r
/
it nov follows from
N(r)
I/Pl rn+
(3.6)
o(1)
o
0(1) t(r Pl ).
Since
p
<
p and p is
definition of
-order.
the
-order of f,
n
sufficiently large p and
(I +
1/(np+
o(I))(- dk"
Further, we have
dk+n p
above
inequality
Thus, equality must hold in (3.4).
PROOF OF THEOREM 2.
a(np+ k))
the
Since
D
2knp+l-np
Pf
are univalent in
contradicts
This proves the lemua.
A, from (2.2), we get for
+ I.
k)
.d
d
the
(3.8>
1/(np+k)
P
l=2{(nl-ni_l+l)d(ni-nl_l+l)--.d(2)
1/(np+k)
GROrH OF ENTIRE FUNCTIONS WITH UNIVALENT DERIVATIVES
ll(n +k)
P
(d
(dk...d
25
11 np+
(np+1 -rip+ I) ...d
and
"(dnP +k...d I)-I/(rip+k)
(d(n +2)..,d(1))
P
(3.8),
Using these inequalities, (2.5) and
a
-I/(rip+2).
it follows that, for sufficiently large p,
1/(np+k)
(np+k)
(3.9)
p+l
-nl-ni-1 )/n P
i=2 (d(nl-ni-1 +I)
2(1+o(I))
I/n p
d(n )...d(1))
P
M
Since log
{log d( nl-ni_ +I ):
max
P
d(n) is the restriction of a slowly oscillating function on integers, by
(3.1)
p+l
i2 d(nl-n i-I +I) (nl-nl_)/n p
(d(np)...d(1)) 1/np
log
n
p+l
_InP [iZ=2
n
P
(ni-ni-I)
d(nl-ni_
log
+)
og d(1)
-iz=
Mp+l -log d(np)
Consequently, for sufficiently large p,
(n
+k) log d(n
_p.
P
log
+k)
og
-ogl.Cnp+k)
d(np+1+1)
n
d(np)
Again, from the definition of S.O. function
P
log
Mp+l -log
d(np)
log
2
d(np+
as
p+(R).
l-lent e,
lira
(3.t0)
M
sup__ p,,
p +
log d(n
P
M is bounded, there is nothing to prove.
p
For p > 2, let,
If
A
So, let
M
p
as
log d(n -np-I +1)
p
p
and
iog d(n p
M
B
But as
M
P
"---P
p
d(np)
d(nl-ni_1+l):
log
max {log
2lp}, for each p
)
2, there is some
26
qp’ qp
B
KAPOOR
G.P.
A
P
qp qp- 1+1).
-n
8ence
qp
Taking
qp
log d(n
M
p
p such that
O.P. JUNEJA AND J. PATEL
llm sup A
P
B
lira sup
p/(R)
P
p+
Now (3.2) follows from (3.10).
COROLLARY.
Suppose the conditions of Theorem 2 are satisfied.
d(np-np_ I)
log
Le__t
THEOREM 3.
Dnh
is univalent in
PROOF.
Suppose
I.
p
0
non-negatlve integers.
o(log
>
Le__t
{dn}n.
0 and
p
/
",
d(np))
{np}v.
I_
be a strictly increasing sequence of
Then, there is an entire function h of
A if and only if n-n_ for some p.
p
If as
V-order p such that
is an increasing sequence of positive numbers
such that log d(n) is the restriction of a slowly oscillatlng function on integers and
dl=l.
Let,
2Pd(np+l)...d(2)
hi+l=
if
(J-n +1
J=nP
0
Define,
h(z)=
Zj. 0
o(h)
k
/
otherwise.
Then, h(z) is an entire function and
h z
lira sup
for some p
P
k log
log
(n +I) log d(n +I)
P
P
llm sup
p
p log 2 +
log(d(n_+l).,,d(2))
To show that
D
nph
"
given by
n
D
Ph(z)
is univalent in
Zk=0 (np+k-np+l)
A,
P
d(n_., + )...d 2
.d(2)
d(np+-np+l)P’r"
it is enough to prove that
d(np+k+1) ...d(2)
r’kf1(np+k-np+l) d(np+k-np+1).. *d(2) [h(np+k+1)l
d(np+l)...d(2)lh(np+1)[.
h(np+k+l)z
np+k-np+
GROWTH OF ENTIRE FUNCTIONS WITH UNIVALENT DERIVATIVES
Since
27
I,
p
d
r’k. (np+k-np+ 1)
(np+k+l)
.d( 2
d(np+k-n/1)...d(2 Ih(np+k+l)l
(d(np+k+l)...d(2))
d(np+ k- n P+l)...d(2)
< _I
2p
Id(np+1)
.d(2))
k=1
2-
d(np+l)...d(2) h(np+l) I.
As
n
Dn+lh(o)
If
o=O
0 unless
nfn
then take
h+
hi+ 1
P
for some p, only
D
Ph
are univalent in
defined by
if
2P+d(np+I)’" "d(2) (J-np+l)
J=nP
0
in place of
hi+
A.
for some p.
otherwise
in the Taylor series of the function h(z).
REFERENCES
I.
3.
GELFOND, A.O. and LEONTEV, A.F. On a generalization of Fourier series, Mat. Sb.
(N.S.) 29 (71), (1951), 477-500.
BOAS, R.P., BUCK, R.C. Polynomial expansions of analytic functlons Ergebnlsse
der Mathematik und lhrer Grenzgeblete, 19, Sprlnger-Verlag, 1964.
BOAS, R.P. Entire Functions, Academic Press,. 1954.
4.
SHAH, S.M. and TRIMBLE, S.Y.
2.
The order of an entire function with some
derivatives univalent, J. Math. Anal. Appl. 46
5.
SHAH, S.M.
(1974), 395-409.
Analytic functions with some derivatives univalent and a related
conjecture, Attl. Acad. Naz Lincei Rend CI. Sci. Fis. Mat. Natur. 61 fasc. 5
(1976), 344-353.
6. BRANGES, L.De. A proof of Biberbach Conjecture, Acta. Math. 154 (1985), 137-152.
7. SHAH, S.M. and TRIMBLE, S.Y.
Entire functions with some derivatives univalent,
Canadian J. Math. 26 (1974), 207-213.
28
8.
9.
G.P. KAPOOR, O.P. JUNEJA AND J. PATEL
JUNEJA, O.P. and SHAH, S.M. Univalent functions with univalent Gelfond-Leontev
derivatives, Bull. Austr. Math. Soc. 29 (1984), 329-348.
Note on Fourier serles (III), Asymptotic
HARDY, G.H. and ROGOSINSKI, W.W.
formulae for the sums
of certain trigonometrical ser’les,
(Oxford Set.), 16 (1954), 49-58.
.uart.
J.
Math.