Internat. J. Math. & Math. Sci.
Vol. 9 No. (1986) 197-200
197
A NOTE ON SUBORDINATION
SHIGEYOSHI OWA
Department of MathematicsKinki University Osaka, Japan (Received March 20, 1984)
ABSTRACT. Suffridge showed a result for subordinate functions. The object of the pre- sent paper is to show some subordinate theorems with the aid of the result by Suffridge.
}’,’}Y WORD[; A?JD
P!!"ASEL’.
Suboz,dinate,
Starlikeof
order,
convexof order
1980 AMC SU/JECT CLASSIFfCATTONCODE.
ZOC45.1. INTRODUCTION.
Let f(z) and g(z) be analytic in the unit disk {z:
[z[
1}. A function f(z) is said to be subordinate tog(z)
if there exists a function(z)
analytic in the unit disk u satisfying (0) 0 and[(z)[
(z such that f(z) g(+(z)) for z e We denote byf(z)’g(z)
this relation. In particular, if g(z) is univalent the unit disk u the subordination is equivalent to f(O) g(O) and range f(z) range g(z).This concept of subordination can be traced to LindelBf
[I],
but Littlewood[2],[3]
and Rogosinski
[_4],
5] introduced the term and discovered thebasic properties. Recently Suffridge 6] and Hallenbeck and Ruscheweyh [7] studied the subordinate functions and showed many interesting results for subordinations.Let A denote the class ,f functions of the form
f(z) z
+
a z (1 1)n=2 n
which are ana|ytic in the unit disk u. Further let S be the subclass of A con-
sistin
of ana|vtic and univalent functions in the unit disku.
Then a functionf(z)
of S is said to be starlike of order c if and only if (z)
t
R
YzT-
(z u)for some (0
__<_ ,
I) We depote byS*()
the class of all starlike functions of order o Further a function f(z) of S is said to be convex of order if andonl
ifzf"(z)
Re
+
a (z u)f’ (z)
fr some r (0
__
1). And we denote bv K(a the class of convex functions of198 S. OWA
order a. it is well-known that f(z) K(a) if and only if
zf’(z)
S*
(a), S (a) S K(a) K, and S (0) S K(O) K for a O.The classes S (a) and K(a) were first introduced by Robertson
[8],
and latter studied by Schild [9], MacCregr[I0]
and Pinchuk[II].
Further, recently,some]
,
classes defined by using the extremal function z/(l-z)2(l-a)
for S (a) were studied by Ruscheweyh
[12],
Sheil-Small, Silverman and Silvia[13],
Silverman andSilvia [14], and Ahuja and Silverman
[15].
Our main tool in this paper is the following result by Suffridge
[15].
Zn
LEMMA. Let the function f(z)
n
2an
be analytic in the unit disk u andthe function g(z) be in the class
S*
If f(z) is subordinate to g(z), that is,(z)-g(z),
theniz
f(t) dt< I
zg(t)
dt0 t 0 t
for z
(r) z: Izl
r, 0 r 1}.2. SUBORDINATION THEOREMS.
In this section, we show some subordination theorems with the aid of Lemma.
THEOREM i. Let the function f(z) defined by (1.1) be in the class of K(a).
"0
Then f’(reI (0 r I) is contained in the image domain of the closed disk u(r)
4(a-l)/(l-z)
under the function e Further it lies for r
#
0 on the boundary of of this image domain if and only iff(z)
,:z e4(l-a)/(1-et)dt.
(2.1)0 where
PROOF. Since f(z) is in the class K(a), f(z) satisfies that Re
zf"(z)
> a-I (z u)
f’(z)
Note that zf"(z)/f’ (z)
2a2z +
is analytic in the unit disk H, and that thefunction
z/(l-z)
2 is starlike with respect to the orgin and Re{z/(l-z) 2}
-I/4.Hence we have that
f"(z)
4(l-a)zz
(z)(1_z)e
(Z 6 U).Consequently, by using Lemma, it follows that log
f’(re
is contained in the image domain of u(r) under the function4(a-l)/(l-z),
where log is understood to be that branch which vanishes at the point one. Thus we can see that f’(rei8)
lies for4(a-l) (l-z). Further r
#
0 on the boundary of the image domain of u(r) under ef’(re
io)
lies for r 0 on the boundary of the image domain u(r) under 4(a-l) (l-z)e if and only if
zf"(z)
4(l-a)ez (ll
f’k:’)
(1_Z)2
hence further f(z) i,- the function of the form (2,1), This completes the proof }f the theorem.
NOTE ON SUBORDINATION 199 THEOREM 2. Let the function f(z) defined by (l.I) be in the class
S*(=).
Then f(re
i0)/re
i0 (0__<
r < I) is contained in the image domain of the closed disk 4(-I) (l-z)u(r) under the function e Further it lies for
r#
0 on the boundary 4(l-a)/
(l-ez)of this image domain if and only if f(z) ze where
I[
I.PROOF. Since f(z) e
S*(a),
f(z) satisfies that Rezf’(z)
> a(z
u) f(z)and the function zf’
(z)/f(z) -I a2z +
is analytic in the unit disku.
Hencezf’
(z)/f(z)
takes values in the image domain of the unit disk u under the func- tion4(l-a)z/(l-z)
2 that iszf’
(z) 4(l-a)zf(z)
I<
(l-z)2 (z e U).By
virtue of Lemma, we observe that logf(rei)/re
I (0 r < I) is contanined in in the image domain of u(r) under4(a-l)/(l-z)
and it lies for r#
0 on the boundary of this image domain if and only ifzf’(z) 4(1-a)ez
(lel
I)f(z) (l_ez)2
4(l-a)
(1-ez)
hence further, f(z) ze This gives the result we require.
Finally we show a theorem for functions f(z) satisfying
Re{zf’(z)}
a(a0).
THEOREM 3. Let the function f(z) defined by
(LI)
satisfyRe{zf’(z)}
> a(a>O).
Then f(re
io)
(0__<
r < 1) is contained in the image domain of the closed disk u(r) under the function-4a/(1-z).
Further it lies for r 0 on the boundary of this do- main if and only if f(z)=-4a/(l-z),
wherele =I.
PROOF. We note that the function
zf’(z)
z+ 2a2z2 +
takes values in theimage domain of the unit disk
u
under the function-4az/(l-z)
2 which belongs to the classS*.
Therefore we’can prove the theorem by using the same technique as in the one of Theorem with the aid ofLemma.
REPERENCES
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J.E. On inequalities in the theory of functions,Proc.
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J.E. Lectures on the theory of functions, OxfordUniversit Press,
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ROGOSINSKI, W.
On subordinate functions,Proc. Cambrid
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W. On the coefficients of subordinate functions,Proc.
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T.J. Some remarks on convex maps of the unit disk, Duke Math. J. 37(1970), 775-777.
200 S. OWA
7. HALLENBECK,
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(1975), 191-195.
8.
ROBERTSON,
M.S. On the theory of univalent functions, Ann. Math. 37(1936),
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MacGREGOR,
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SILVERMAN,
H. andSILVIA,
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AHUJA,
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