[r, ternat. J. Math. & Math. Sci.
Vol. 8 No. 4 (1985) 677-681
677
ON CERTAIN INEQUALITIES FOR SOME
REGULAR FUNCTIONS IN Izl<l
MILUTIN
OBRADOVI
Department of Mathematics
Faculty of Technology and Metallurgy
University of Belgrade, Yugoslavia
(Received August 6, 1985)
ABSTRACT.
z+a2z
2
In this paper we give some inequalities for regular functions f(z)
+... in
Iz <I
especially for starlike and convex functions of order
C<I. To some extent those inequalities are the generalisations and improvements
of the previous results given by Bernardi
111
Some interesting consequences are
given, too.
KEY WORDS AND PHRASES.
!i(gula, fun:tion.?, Starlike
functions of
orler
,
Convex
INTRODUCTION
Let A denote the class of functions f(z)=
unit disc E-
z+a2z
2
+... which are regular in the
{z’Iz!<l}.
For a function f_A we say that it is starlike of order c, 0<I, in E, if
and only
Re
zf (z) >, z:E.
f(z)
The class of such functions we denote by
S*
(1.1)
S*(). It
is evident that
S*() S*,
where
is the class of starlike functions in E.
Similarly, for a function fA we say that it is convex of order c, Oc(I, in
E, if and only if
Re{ 1+
zff(z)(z)
>
(1 .2)
zE E.
The class of such functions we denote by K(a). In this case we have also that
K() K(O)--K, where K
that
S*
is the class of convex functions in E.
It is well-known
and K, are the subclasses of the class S of univalent functions in E [2]
In this paper we give some inequalities of the integral type for the functions of A, especially for the functions of the classes S*() and K(), 0(I. For
special cases of those inequalities we have some improved inequalities of Bernardi
In this sense, our inequalities can be considered as the general isations and
I]
improvements of the corresponding inequalities of Bernardi.
For the proofs of the coming inequalities we will use the following result
of Miller [3]
THEOREM A.
domain in C x
C),
Let (u,v) be a complex function
and let u=
u1+iu2,
v=
D-C (C-complex plane, DSuppose
that the function
sarisv1+iv 2.
"
OBRADOVI
M.
678
ed the next conditions:
(a) (u,v) is contiuous in D;
(b)
(I,0) D and Re(l,0)
(c)
for all
Re(u2i,vl)_<0
0;
2
<-(1/2) (l+u 2 ).
D and such that v
(u2i,v I)
Le’t p(z)= |+plz+... be regular in the unit disc E such that (p(z),zp ’(z))ZD
for all zE. If
zEE,
Re(p(z), zp"(z)) >0
then
Rep(z)>O,
zE E.
INEQUALITIES AND CONSEQUENCES
Let fE A,
THEOREM I.
<
(z)
>=Re
Re
and a>-1. Then the following implication
z
a+1
c
ta-lf(t)dt>a+3---,
f
Z
O
z.E,
(2.1)
(1-[3)p(z).
(2.2)
is true.
PROOF.
Let’s put
1-
/zo
a+l
=IS and
’+"a
Z
ta_lf(t)dt= 6+
p(O)=l. From (2.2)
Then it is easily shown that 6<I, p is regular in E and
we ha-
ve that
z
a+l
a
f
f(t)dt
O
Z
a---T- 6+(1-)p(z)]
and hence, after differentiation and some simple transformations, we have that
f(Z)z
Since Re
f
(z-----)>,
Z
z E,
-
-a=6
+(1-6)P(Z)+
Re{6-+ (1-6)p(z)
-
(u,v)
(2.3)
zp (z).
(2.3) we get
then from
Now we may consider the function
]-6
+
1-6 zp’(z)}>O,
+-1-6
(1-[3)u+i-
(2.4)
zEE.
v
sati(it is noted u=p(z), v=zp’(z)). It is directly checked that the function
(a), (b) and (c) of Theorem A. Namely, in this case it is
2
is continuous in D (I O) D Re(1 O)= I->0 while for all
D= C
sfies the conditions
(u2i,vl)D
v1<-(I/2)-(1+u )
such that
Re(u2i,v I)
B
-
we have that
+ I-3
T Vl <6-+
I-
2
I-(I+u2)]
-2
I-[3
a+1
u22< O
Therefore, by applying Theorem A we hav that Rep(z)>O, zE; hence using (2.2)
we fina!ly get
Re
z
a+1
Z
f
a-lf(t)dt>6,
z E,
O
which was to be proved.
COROLLARY I.
If
f:S*(),
Re
<<I,
f(z) >
then it is known
3-2c
zE,
14] that
679
INEQUALITIES FOR SOME REGULAR FUNCTIONS
for those classes from Theorem
and
we have that
a+l
Redo ta-lf(t)dt>
Because f K(():
+2
zf’(z)IE S*((),
COROLLARY 2.
If fie
f----) >
f(z)
2(1_21-2)
Z
rz ta-lf(t)dt>3-21l()
o
Especially, for =O, i.e. for fK, we
z
a+1
z
Re
then from 15] we have
(2.7)
2
and from Theorem
+_
(2.6)
2
21og2
a+_ _ 1_
5-2c
3(3_2a)
--
2-
Re
2-<
>
z
whe re
1(,)
I__<
K()
K(c), O<<l,
Re
(2.5)
zE E.
0< <I, is true, then from (2.5) for a=O
we get the following estimates for f
Re
-I-2) (3’+2’a)
a-1
a+- Jo
] (6)
+ 2
.(3.=2131(a))(3+2a
have that I(O)= and
2+a
f(t)dt>3+2-
zE
from (2.8)"
(2.9)
zIE E,
which improves the earlier result of Bernardi !11,
(2.8)
Th.8(A), where
the constant or
(2.9) is equal to 1/2 (namely, 2+a
for a>-1).
If f A and zf" (z). K(c), O< I, then from (2.8) for a=O we have also that
5 2 ()
f z
>
Re
(2.10)
z(E,
z
->
the right side of
3.(..3_2131((})
where L () is defined by (2.7).
If we put
COROLLARY 3.
Ref
zf’(z)
a+1 z
(z)>a=Re ----/
a
instead of fEA we get that
I-(
(t)dt>+3--a
f
o
z
s
in Theorem
(2.11)
z_E,
rue.
From (2.|I) for a= O we have that
Ref (z)>= Re
f(z)
T>
is true. From (2.12) by applying Theorem
2+1
3
(2.12)
zE,
once again we have that the following
impl cat ion
Ref
(z)>
z
a+1
ta_if(t)dt>2+ +2_. 3+2a’IRe--T
z
z:E,
(2.13)
is true for f(-A, <I and a>-1.
Let f
THEOREM 2.
Re
z
z
#
o
S*(),
a f(z)
a-1
f(t)dt
O<<I, and let a>max{-1,-2c}, then we have that
>
2a+2(_
1+V/2a+2_ 2+8 (a+l)
4
ziEE
(2 14)
OBRADOVI
M.
680
PROOF.
The inequality
(2.14) is equivalent to the inequality
a
z f(z)
z
Re
> 2a+2cz-
+x/( 2a+2- 2+8 (a+
(a+1)# a-I f(t)dt
zEE
4 (a+1")
(2
14")
o
Let’s put
zaf(z)
z
(a+l)f
a-1
+
(2.15)
(1-)p(z),
f (t)dt
o
(2.14).
It is easy to
where B is the number or the right side of the inequality
it
that the funcbe
shown
can
E
(moreover,
in
is
0<-:;<
p(z)
I,
regular
show that
z
a-1
a+1
tion
#t ’f(t)dt S*(c) if f S*(), Or<l, a:>-l. Hence, the function p(z)
a o
z
may have the removable singularity in z=O) and p(O)= I. From (2.15) after logarithmic differentiation we may get
zf’ (z)
f(z)
zf’(z) >
(Z)
Re
and since
-=
zp’ (z)
(a+l) -a-+(a+l) (1-F) p (z) +(1-l) +(1-1g)
p(z)
,
z E,
then
Ret(a+l)8-a-+(a+l)(1-g)p(z)+(1-1g)
zp’(z)
g+(Z)p(z)}>O,
z(E.
(2.16)
In this case we consider the function
-
(u,v)
The function
also that
(a+l)ig- a-o+ (a+l) (1-t)u+(1-)
C, where D= ({-
D
(I,0) D, Re(1,0)= l-c>O,
v1<-(I/2)(1+u22
-_R})x
C
v
and
while for all
(2.17)
+(’l-B)u
is continuous in D. We have
(u2i,vl).
D so that
we get
v
Re{
(u2i,v I)
(a+l)ig-a-a+(l-6)g
< (a+l)-a-c+(-)
B2+(1-13)2u
132+(I
B)2u22
-(1/2) (l+u 2)
2
B2+(l_g)2 u 22
l+2(a+)-2(a+l)Blu 22
(we used the fact that lg is the solution of the equation 2(a+l)B2-(2a+2-l)g-1=O).
To prove that Re(u2u,vl)<O we must show that 1+2(a+)-2(a+l)EO, what is not dif-
,
ficult regardin 9 the values for
and B. Therefore, from Theorem A and (2.16) we
have that Rep(z)>O, zE, which is equivalent to (2.14"), i.e. (2.14).
COROLLARY 4. If fK(), 0<<I, what is equivalent to
from (2.14) for a=O we have that
Re
z_f’(z)>2c
f(Z)
I+2/’
1) 2 +[3
This is the former result given by Jack 16].
z
E.
zf’(z)eZS*(),
then
(2.18)
681
INEQUALITIES FOR SOME REGULAR FUNCTIONS
,
If f K then Re zf’(z)>1 [2], (which also implies from (2.18)
COROLLARY 5
f(z)
=O) and because of that, from Theorem 2 we have
for
Re
zaf(z)
a+
>
z
v/ a 2 + 2a+ 2
which improves the former result of Bernardi [I],
(2.19)
the right side of
in [I]
+2a+2
a+
2
THEOREM 3.
-
Let f
z fl
Re
r(z)
a
Z
PROOF.
tion
zf(z)
is
,
I+2a
>---2--
K(),
I+2a
a+I)2 +I
(=
>
Th.7(B). Namely, the
constant on
but
O<,z< l, and
(z)
(2.19)
zE,
2
# taIf(t)dt
o
> a+1).
a>max{-1,-2}, then we have
2a+2-|
+
/(2a+2-I
4
jz ta_If(t)dt
o
2+8(a+I) z
(2.20)
E.
If f K() then zf(z) S*(), and by applying Theorem 2 to the funcand partial integration in corresponding integral we have the state-
3.
COROLLARY 6. For
ment of Theorem
c=
O, i.e. for f. K, we have that
zf’ (z)
Re
f(z)
a
Z
> 2a-
+x/a-l)2+8(a+l)
/ta-lf(t)dt
4
which also improves the result of Bernardi [I],
z
E
Th.8(C).
REFERENCES
BERNARDI, S.D. Convex and starlike univalent functions, Trans.Amer.Math.
Soc. Vol. 135(1969), 429-446.
POMMERENKE, C.
MILLER, S.S.
Univalent functions, Vandenhoeck & Ruprecht, Gtingen (1975).
Differential
inequalities and Caratheodory functions, Bull.
._A_m_e_r. _Math. So_c:__81(I)(1975), 79-81.
Estimates of the real part of f(z)/z for some classes of univalent functions, Mat. Vesnik 36(4)(1984), 226-270.
4.
OBRADOVIC], M.
5.
OBRADOVI(, M., OWA, S.
6.
,
Mat.
On some results of convex functions of order
Vesnik, to appear.
JACK, I.S. Functions starlike and convex of order c, J.London Math.Soc.
3(2)(1971), 469-474.