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Differential subordination and superordination-preserving integral operators

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Abstract

In the first section we consider the function φ:ℂ 2 →ℂ analytic in a domain Δ⊂ℂ 2 . Let p be an analytic function in the unit disk U such that φ(p(z),zp ' (z)) is univalent in U and suppose that p satisfies the so called generalized Briot-Bouquet differential superordination h(z)≺φ(p(z),zp ' (z))=α(p(z))+β(p(z))γ(zp ' (z)). We determine conditions on h,α,β and γ so that the above subordination implies q(z)≺p(z), where q is the largest function so that q(z)≺p(z) for all p satisfying this first-order differential superordination. In the second section, for the integral operator A β,γ :K→H(U) defined by A β,γ (f)(z)=β+γ z γ ∫ 0 z f β (t)t γ-1 dt 1/β ,β,γ∈C, we determined sufficient conditions on g,β and γ such that zg(z) z β ≺zf(z) z β ⇒zA β,γ (g)(z) z β ≺zA β,γ (f)(z) z β · We prove that zA β,γ (g)(z) z β is the largest function so that the right-hand side of the above implication holds, for all f functions satisfying the left-hand side differential superordination. The concept of differential superordination was introduced by S. S. Miller and P. T. Mocanu [Complex Variables, Theory Appl. 48, No. 10, 815–826 (2003; Zbl 1039.30011)] like a dual problem of differential subordination (see S. S. Miller and P. T. Mocanu [Differential subordinations: theory and applications. Pure and Applied Mathematics, Marcel Dekker. 225. New York, NY: Marcel Dekker. xi, 459 p. (2000; Zbl 0954.34003)].
UDC 532.595
Differential subordination and
superordination-preserving integral
operators
T. Bulboac˘a
(Faculty of Mathematics and Computer Science, Babs-Bolyai University,
Cluj-Napoca, Romania)
bulboaca@math.ubbcluj.ro
In the first section, consider the function ϕ:C2Canalytic in a domain
C2, let pan analytic function in the unit disc Usuch that ϕ(p(z), zp0(z)) is
univalent in Uand suppose that psatisfies the so called generalized Briot-Bouquet
differential superordination
h(z)ϕ(p(z), zp0(z)) = α(p(z)) + β(p(z)) γ(zp0(z)).
We determined conditions on h,α,βand γso that the above subordination
implies q(z)p(z), where qis the largest function so that q(z)p(z)for all p
functions satisfying this first-order differential superordination.
In the second section, for the integral operator Aβ,γ :K → H(U), with K ⊂ H(U),
defined by
Aβ,γ (f)(z) = β+γ
zγZz
0
fβ(t)tγ1dt1
, β, γ C,
we determined sufficient conditions on g,βand γsuch that
zg(z)
zβ
zf(z)
zβ
zAβ,γ (g)(z)
zβ
zAβ,γ (f)(z)
zβ
.
We proved that zAβ,γ (g)(z)
zβ
is the largest function so that the right-hand
side of the above implication holds, for all ffunctions satisfying the left-hand
side differential superordination.
The concept of differential superordination was introduced by S. S. Miller and P.
T. Mocanu in [8] like a dual problem of differential subordination [7].
c
T. Bulboac˘a, 2004
2T. Bulboac˘a
1. Generalized Briot-Bouquet superordinations. Let denote by
H(U) the class of analytical functions in the unit disc U = {zC:|z|<1}
and for aCand nNlet
H[a, n] = fH(U) : f(z) = a+anzn+an+1zn+1 +. . . , z U.
If f, F H(U) and Fis univalent in Uwe say that the function f
is subordinate to F, or Fis superordinate to f, written f(z)F(z), if
f(0) = F(0) and f(U) F(U).
Let ϕ:C3×UC, let hH(U) and qH[a, n]. In [8] the authors
determined conditions on ϕsuch that
h(z)ϕ(p(z), zp0(z), z2p00(z); z)
implies q(z)p(z), for all pfunctions that satisfies the above
superordination. Moreover, they found sufficient conditions so that the
qfunction is the largest function with this property, called the best
subordinant of this subordination.
The present paper deals with the case when
ϕ(p(z), zp0(z)) = α(p(z)) + β(p(z)) γ(zp0(z)),(1)
hence we study the so called generalized Briot-Bouquet differential
superordination. We determine conditions on h,α,βand γso that this
subordination implies q(z)p(z), and we find its best subordinant q.
Note that the first type of differential superordinations represents a
generalization of the Briot-Bouquet differential superordination, i.e.
h(z)p(z) + zp0(z)
βp(z) + γ,
obtained from (1) for α(w) = w,β(w) = 1
βw +γand γ(w) = w.
Definition 1 [8] We denote by Qthe set of functions hthat are analytic
and injective on U\E(h), where
E(h) = ζU : lim
zζh(z) = ,
and such that h0(ζ)6= 0 for ζU\E(h).
Differential subordination and superordination-preserving ... 3
Theorem 1 [2, Theorem 3.1.] Let qbe a convex (univalent) function in
the unit disc U, let α,βH(D), where Dq(U) is a domain, and let
γH(C). Suppose that
Re α0(q(z)) + β0(q(z)) γ(tzq0(z))
β(q(z)) γ0(tzq0(z)) >0,zUand t0.
If pH[q(0),1] ∩ Q, with p(U) D, and α(p(z)) + β(p(z)) γ(zp0(z)) is
univalent in U, then
α(q(z)) + β(q(z)) γ(zq0(z)) = h(z)α(p(z)) + β(p(z)) γ(zp0(z))
q(z)p(z),
and qis the best subordinant.
Taking β(w)=1in the above theorem we get the next corollary:
Corollary 1 Let qbe a convex (univalent) function in the unit disc U,
αH(D), where Dq(U) is a domain, and let γH(C). Suppose that
Re α0(q(z))
γ0(tzq0(z)) >0,zUand t0.
If pH[q(0),1] Q, with p(U) D, and α(p(z)) + γ(zp0(z)) is univalent
in U, then
α(q(z)) + γ(zq0(z)) α(p(z)) + γ(zp0(z)) q(z)p(z),
and qis the best subordinant.
For the particular case when γ(w) = w, using a similar proof as in
Theorem 1 (see [2]) we obtain:
Corollary 2 Let qbe a univalent function in the unit disc Uand let α,
βH(D), where Dq(U) is a domain. Suppose that
(i) Re α0(q(z))
β(q(z)) >0,zU,
(ii)Q(z) = zq0(z)β(q(z)) is a starlike (univalent) function in U.
If pH[q(0),1] ∩ Q, with p(U) D, and α(p(z)) + zp0(z)β(p(z)) is
univalent in U, then
α(q(z)) + zq0(z)β(q(z)) α(p(z)) + zp0(z)β(p(z)) q(z)p(z),
and qis the best subordinant.
4T. Bulboac˘a
For the case β(w) = 1 and using the fact that the function Q(z) =
zq0(z)is starlike (univalent) in Uif and only if qis convex (univalent) in
U, Corollary 2 becomes:
Corollary 3 Let qbe a convex (univalent) function in the unit disc Uand
let αH(D), where Dq(U) is a domain. Suppose that
Re α0(q(z)) >0,zU.(2)
If pH[q(0),1] ∩ Q, with p(U) D, and α(p(z)) + zp0(z)is univalent in
U, then
α(q(z)) + zq0(z)α(p(z)) + zp0(z)q(z)p(z),
and qis the best subordinant.
Particular cases. Next we will give some particular cases of the above
results obtained for appropriate choices of the q,αand βfunctions.
Taking α(w) = wand β(w) = 1,Re γ > 0, in Corollary 2, condition
(i)holds if Re γ > 0and (ii)holds if and only if qis a convex (univalent)
function in U, hence we obtain:
Example 1 [8, Theorem 8] Let qbe a convex (univalent) function in the
unit disc Uand let γC, with Re γ > 0. If pH[q(0),1] ∩ Q and
p(z) + zp0(z)
γis univalent in U, then
q(z) + zq0(z)
γp(z) + zp0(z)
γq(z)p(z),
and qis the best subordinant.
Considering in Corollary 3 the case α(w) = ew, then condition (2)
becomes
Re α0(q(z)) = eRe q(z)cos(Im q(z)) >0, z U,
that is equivalent with (3), and we get the following result:
Example 2 Let qbe a convex (univalent) function in the unit disc Uand
suppose that
|Im q(z)|<π
2, z U.(3)
Differential subordination and superordination-preserving ... 5
If pH[q(0),1] ∩ Q and ep(z)+zp0(z)is univalent in U, then
eq(z)+zq0(z)ep(z)+zp0(z)q(z)p(z),
and qis the best subordinant.
Remark 1 Taking q(z) = λz,|λ| ≤ π/2in Example 2 we have the next
result:
If pH[0,1]∩Q such that ep(z)+zp0(z)is univalent in Uand |λ| ≤ π/2,
then
eλz +λz ep(z)+zp0(z)λz p(z),
and λz is the best subordinant.
If we consider in Corollary 3 the case α(w) = w2
2βw, then we may
easily see that (2) is equivalent with (4), and we deduce the next example:
Example 3 Let qbe a convex (univalent) function in the unit disc Uand
suppose that
Re q(z)> β, z U.(4)
If pH[q(0),1] ∩ Q and p2(z)
2βp(z) + zp0(z)is univalent in U, then
q2(z)
2βq(z) + zq0(z)p2(z)
2βp(z) + zp0(z)q(z)p(z),
and qis the best subordinant.
Remarks 2 1. The function q(z) = eλz is convex (univalent) in Uif
and only if |λ| ≤ 1, and under this assumption Re q(z)>0,zU.
Using the result of Example 3 for this choice we obtain:
If pH[1,1] ∩ Q such that p2(z)
2+zp0(z)is univalent in Uand
|λ| ≤ 1, then
e2λz
2+λzeλz p2(z)
2+zp0(z)eλz p(z),
and eλz is the best subordinant.
6T. Bulboac˘a
2. The function q(z) = 1 + (2β1)z
1 + z,β < 1, is convex (univalent) in
Uand Re q(z)> β,zU. Hence, by using Example 3 we have:
If pH[1,1] Q such that p2(z)
2βp(z) + zp0(z)is univalent in U
and β < 1, then
12β2(2β24β+ 3)z+ (1 2β)z2
2(1 + z)2p2(z)
2βp(z) + zp0(z)
1 + (2β1)z
1 + zp(z),
and 1 + (2β1)z
1 + zis the best subordinant.
Remark 3 Another study concerning different generalizations of the
Briot-Bouquet differential subordinations and superordinations, together
with the correspondent sandwich-type theorems and interesting particular
cases, may be found in [5].
2. Superordination-preserving integral operators. Let the integral
operator Aβ,γ :K H(U),K ⊂ H(U), be defined by
Aβ,γ (f)(z) = β+γ
zγZz
0
fβ(t)tγ1dt1
, β, γ C.(5)
In [1] the author determined conditions on the gfunction and on
parameters βand γso that
zf(z)
zβ
zg(z)
zβ
implies zAβ,γ (f)(z)
zβ
zAβ,γ (g)(z)
zβ
,
and this result that was further improved in [3].
Now we studied the reverse problem, in the sense to determine sufficient
conditions on g,βand γsuch that the next superordination holds:
zg(z)
zβ
zf(z)
zβ
implies zAβ,γ (g)(z)
zβ
zAβ,γ (f)(z)
zβ
,
and we proved that, under our assumptions, this result is sharp.
Differential subordination and superordination-preserving ... 7
Let cCwith Re c > 0and let N=N(c) = |c|1 + 2 Re c+ Im c
Re c. If
kis the univalent function k(z) = 2Nz
1z2, then we define the open door
function Rcby
Rc(z) = kz+b
1 + bz , z U,(6)
where b=K1(c). Note that Rcis univalent in U,Rc(0) = cand
Rc(U) = k(U) = C\ {wC: Re w= 0,|Im w| ≥ N}.
Let Abe the set of those ffunctions that are analytic in the unit disk
Uand normalized by the conditions f(0) = f0(0) 1 = 0.
Lemma 1 [3, Lemma 3.1] Let β, γ C, with β6= 0,Re(β+γ)>0, and
let hA, with h(z)h0(z)/z 6= 0,zU. If fAand
βzf 0(z)
f(z)+ (γ1)zh0(z)
h(z)+1+zh00(z)
h0(z)Rβ+γ(z)
then
FA, F(z)
z6= 0, z Uand Re βz F 0(z)
F(z)+γzh0(z)
h(z)>0, z U,
where
F(z) = Ih;β,γ (z) = β+γ
hγ(z)Zz
0
fβ(t)hγ1(t)h0(t) d t1
and Rβ+γis given by (6). (All powers in Ih;β,γ are the principal ones.)
Note that a more general form of this lemma may be found in [6].
If we denote by
Fβ,γ =fA:βzf0(z)
f(z)+γRβ+γ(z),
and using the fact that Aβ,γ =Iz;β,γ , the next particular case of the above
lemma, obtained for h(z) = z, will be necessary for our results:
8T. Bulboac˘a
Lemma 2 Let β, γ Cwith β6= 0,Re(β+γ)>0. Then f∈ Fβ,γ
implies FA,F(z)
z6= 0, z Uand Re βzF 0(z)
F(z)+γ>0, z U,
where F(z) = Aβ,γ (f)(z).
First we need to determine the subset K ⊂ H(U) such that the integral
operator Aβ,γ will be well-defined. By Lemma 2, if we choose the set
K ≡ Fβ,γ then the integral operator defined by (5) will be well-defined on
K.
Theorem 2 [4, Theorem 3.1.] Let β, γ Cwith β6= 0,1< β +γ2. Let
f, g ∈ Fβ,γ , and for β6= 1 suppose in addition that f(z)/z 6= 0,g(z)/z 6= 0
for zU. Suppose that
Re 1 + zϕ00(z)
ϕ0(z)>1(β+γ)
2, z U,
where ϕ(z) = zg(z)
zβ
.
Let f Q ∩ Fβsuch that zf(z)
zβ
and zAβ,γ (f)(z)
zβ
are
univalent functions in U. Then
zg(z)
zβ
zf(z)
zβ
implies zAβ,γ (g)(z)
zβ
zAβ,γ (f)(z)
zβ
,
and the function zAβ,γ (g)(z)
zβ
is the best subordinant.
If we combine this result together with Theorem 1 of [3], then we obtain
the following differential sandwich-type theorem:
Theorem 3 [4, Theorem 3.2.] Let β, γ Cwith β6= 0,1< β +γ2.
Let f, g1, g2∈ Fβ,γ , and for β6= 1 suppose in addition that f(z)/z 6= 0,
gk(z)/z 6= 0 for zUand k= 1,2. Suppose that the next two conditions
are satisfied
Re 1 + zϕ00
k(z)
ϕ0
k(z)>1(β+γ)
2, z U,for k= 1,2,
Differential subordination and superordination-preserving ... 9
where ϕk(z) = zgk(z)
zβ
and k= 1,2.
Let f Q ∩ Fβsuch that zf(z)
zβ
and zAβ,γ (f)(z)
zβ
are
univalent functions in U. Then
zg1(z)
zβ
zf(z)
zβ
zg2(z)
zβ
implies
zAβ,γ (g1)(z)
zβ
zAβ,γ (f)(z)
zβ
zAβ,γ (g2)(z)
zβ
.
Moreover, the functions zAβ,γ (g1)(z)
zβ
and zAβ,γ (g2)(z)
zβ
are
respectively the best subordinant and the best dominant.
Remark 4 Note that, for the case β= 1, a such kind of result was
obtained in [8, Corollary 6.1], under the assumption that Re γ0and
g1,g2are convex functions.
Because the assumption of the Theorem 2, that the function
zAβ,γ (f)(z)
zβ
needs to be univalent in U, is not so easy to check, in
the next result we will replace this by another condition, which is more
easy to be verified. The detailed proof may be found in [4].
Corollary 4 Let β, γ Cwith β6= 0,1< β +γ2. Let f, g ∈ Fβ,γ ,
and for β6= 1 suppose in addition that f(z)/z 6= 0,g(z)/z 6= 0 for zU.
Suppose that
Re 1 + zϕ00(z)
ϕ0(z)>1(β+γ)
2, z U,where ϕ(z) = zg(z)
zβ
,
and
Re 1 + zψ00 (z)
ψ0(z)>1(β+γ)
2, z U,where ψ(z) = zf(z)
zβ
.
10 T. Bulboac˘a
If f∈ Q ∩ Fβ ,γ , then
zg(z)
zβ
zf(z)
zβ
implies zAβ,γ (g)(z)
zβ
zAβ,γ (f)(z)
zβ
,
and the function zAβ,γ (g)(z)
zβ
is the best subordinant.
References
[1] Bulboac˘a T., Integral operators that preserve the subordination, Bull.
Korean Math. Soc., 1997, 34(4), – P. 627–636.
[2] Bulboac˘a T., Classes of first-order differential superordinations,
Demonstratio Mathematica, 2002, 35(2).– P. 287–292.
[3] Bulboac˘a T., On a class of integral operators that preserve the
subordination, Pure Math. and Appl., 2002, 13(1-2).– P. 87–96.
[4] Bulboac˘a T., A class of superordination-preserving integral operators,
Indagationes Mathematicae, N.S., 2002, 13(3).– P. 301–311.
[5] Bulboac˘a T., Generalized Briot-Bouquet differential subordinations and
superordinations, Rev. Roum. Math. Pures Appl., 2002, 47(5-6).– P. 605–
620.
[6] Miller S. S., Mocanu P. T., Integral operators on certain classes of analytic
functions, Univalent Functions, Fractional Calculus and their Applications,
Halstead Press, J. Wiley & Sons, New York, 1989.– P. 153–166.
[7] Miller S. S., Mocanu P. T., Differential Subordinations. Theory and
Applications. Marcel Dekker, New York, 1999.– 459 p.
[8] Miller S. S., Mocanu P. T., Subordinants of differential superordinations,
Complex Variables, 2003, 48(10).– P. 815–826.
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