# Differential Subordinations Associated with Multiplier Transformations

Authors: Adriana Cătaş; Georgia Irina Oros; Gheorghe Oros
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

The authors introduce new classes of analytic
functions in the open unit disc which are defined by using multiplier transformations.
The properties of these classes will be studied by using techniques involving
the Briot-Bouquet differential subordinations. Also an integral transform is established.


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## Body

## 1. Introduction and Definitions

Letℋ be the class of  analytic functions in the open
unit disc(1.1)U={z∈ℂ:|z|<1}and
let
ℋ[a,n] be  the subclass of ℋ consisting of functions of the form f(z)=a+anzn+an+1zn+1+⋯.
Let 𝒜(p,n) denote the class of functions f(z) normalized by(1.2)f(z)=zp+∑k=p+n∞akzk(p,n∈ℕ:={1,2,3,…})which are analytic in the open
unit disc. In particular, we set(1.3)A(p,1):=Ap,A(1,n):=A(n),A(1,1):=A=A1.
If a function f(z) belongs to the class 𝒜(n), it has the form(1.4)f(z)=z+∑k=n+1∞akzk(n∈ℕ:={1,2,3,…}).
For two functions f(z) given by (1.4) and
for
g(z) given by(1.5)g(z)=z+∑k=n+1∞bkzk(n∈ℕ),the Hadamard product (or
convolution) (f*g)(z) is defined, as usual, by(1.6)(f*g)(z):=z+∑k=n+1∞akbkzk:=(g*f)(z).
If f and g are analytic in U,
we say that f is subordinate to g,
written symbolically as(1.7)f≺gorf(z)≺g(z)(z∈U)if there exists a Schwarz
function w(z) in U which is analytic in U with w(0)=0 and |w(z)|<1 such that f(z)=g(w(z)), z∈U.We consider the following multiplier
transformations.Definition 1.1 (See [1]).
Letf∈𝒜(p,n).
For δ,λ∈ℝ, λ≥0, δ≥0, l≥0,
define the multiplier transformations Ip(δ,λ,l) on 𝒜(p,n) by the following infinite
series:(1.8)Ip(δ,λ,l)f(z):=zp+∑k=p+n∞[p+λ(k−p)+lp+l]δakzk.It follows from (1.8) that(1.9)Ip(0,λ,l)f(z)=f(z),(p+l)Ip(2,λ,l)f(z)=[p(1−λ)+l]Ip(1,λ,l)f(z)+λz(Ip(1,λ,l)f(z))′,Ip(δ1,λ,l)(Ip(δ2,λ,l)f(z))=Ip(δ2,λ,l)(Ip(δ1,λ,l)f(z)).Remark 1.2 (See [1]).
Forp=1, l=0, λ≥0, δ=m, m∈ℕ0, ℕ0=ℕ∪{0},
the operator Dλm:=I1(m,λ,0) was introduced and studied by Al-Oboudi
[2] which
is reduced to the Sălăgean differential operator [3] for λ=1.
The operator Ilm:=I1(m,1,l) was studied recently by Cho and Srivastava
[4] and by Cho and Kim
[5]. The operator Im:=I1(m,1,1) was studied by Uralegaddi and Somanatha
[6], the operator Dλδ:=I1(δ,λ,0) was introduced by Acu and Owa [7] and the operator Ip(m,l):=Ip(m,1,l) was investigated recently by
Sivaprasad Kumar et al. [8].Iff is given by (1.2), then we
have(1.10)Ip(δ,λ,l)f(z)=(f*φp,λ,lδ)(z),where(1.11)φp,λ,lδ(z)=zp+∑k=p+n∞[p+λ(k−p)+lp+l]δzk.
In particular, we set(1.12)I1(δ,λ,l)f(z):=I(δ,λ,l)f(z).
In order to prove our main results, we will make use
of the following lemmas.Lemma 1.3 (See [9]).
For real
or complex numbersa, b,
and
c(c∉ℤ0−:={0,−1,−2,…}),
the following hold:
(1.13)∫01tb−1(1−t)c−b−1(1−tz)−adt=Γ(b)Γ(c−b)Γ(c)⋅2F1(a,b;c;z)(Rec>Reb>0),(1.14)F21(a,b;c;z)=(1−z)−a⋅2F1(a,c−b;c;zz−1),(1.15)F21(a,b;c;z)=2F1(b,a;c;z),(1.16)(b+1)⋅2F1(1,b;b+1;z)=(b+1)+bz⋅2F1(1,b+1;b+2;z),(1.17)F21(a,b;a+b+12;12)=π⋅Γ((a+b+1)/2)Γ((a+1)/2)Γ((b+1)/2).Lemma 1.4 (See [10]).
Letβ,γ∈ℂ, β≠0 and let h be convex in U,
with(1.18)Re[βh(z)+γ]>0(z∈U).
If the function p∈ℋ[h(0),n],
then(1.19)p(z)+zp′(z)βp(z)+γ≺h(z)⟹p(z)≺h(z).Lemma 1.5 (See [11]).
Letμ be a positive measure on the unit interval I=[0,1].
Let g(t,z) be a function analytic in [0,1]×U,
for each t∈I and integrable in t,
for each z∈U and for almost all t∈I.
Suppose also that(1.20)Re{g(t,z)}>0(z∈U;t∈I),g(t,−r) is real for real r and(1.21)Re1g(t,z)≥1g(t,−r)(|z|≤r<1,t∈I).
If(1.22)g(z)=∫Ig(t,z)dμ(t),then(1.23)Re(1g(z))≥1g(−r)(|z|≤r<1).Lemma 1.6 (See [12]).
Letψ(z) be univalent in the unit disc U and let v and φ be analytic in a domain D⊃ψ(U) with φ(w)≠0,
when w∈ψ(U).
Set(1.24)Q(z):=zψ′(z)φ(ψ(z)),h(z):=v(ψ(z))+Q(z).
Suppose that
(1)
Q(z) is starlike in U and(2)
Re(zh′(z)/Q(z))>0 for z∈U.
If q(z) is analytic in U,
with q(0)=ψ(0), q(U)⊂D,
and(1.25)v(q(z))+zq′(z)φ(q(z))≺v(ψ(z))+zψ′(z)φ(ψ(z)),then q(z)≺ψ(z) and ψ(z) is the best dominant.Lemma 1.7 (See [12, Theorem 3.3d]).
Letβ,γ,A∈ℂ,
with Re[β+γ]>0 and let B∈[−1,0] satisfy either(1.26)Re[β(1+AB)+γ(1+B2)]≥|βA+β¯B+B(γ+γ¯)|,when B∈(−1,0],
or(1.27)β(1+A)>0,Re[β(1−A)+2γ]≥0,when B=−1.
If p∈ℋ[1,n] satisfies(1.28)p(z)+zp′(z)βp(z)+γ≺1+Az1+Bz,then(1.29)p(z)≺q(z)=qn(z)≺1+Az1+Bz,where qn is the univalent solution of the differential
equation(1.30)q(z)+nzq′(z)βq(z)+γ=1+Az1+Bz.In addition, the function qn,
is the best (1,n)-dominant and the function qn is given by(1.31)q(z)=β+γβ[k(z)K(z)]β/n−γβ=zK′(z)K(z)=1βg(z)−γβ,where(1.32)k(z)=zexp⁡∫0z(h(t)−1)t−1dt,K(z)=[β+γnzγ/n∫0zkβ/n(t)t(γ/n)−1dt]n/β,and the univalent function g is given by(1.33)g(z)=1n∫01[k(tz)k(z)]β/nt(γ/n)−1dt.Now we define new classes of analytic functions by
using the multiplier transformationsI(m,λ,l) defined by (1.8) as follows.


## 2. Main Results

Definition 2.1.
Let−1≤B<A≤1, λ>0, l≥0, m∈ℕ∪{0}.
A function f∈𝒜(n) is said to be in the class S(m,λ,l;A,B) if it satisfies the following
subordination:(2.1)l+1λ⋅I(m+1,λ,l)f(z)I(m,λ,l)f(z)−1−λ+ll+1≺1+Az1+Bz(z∈U).Remark 2.2.
We note that(2.2)S(0,1,0;1−2α,−1)≡S*(α),S(1,1,0;1−2α,−1)≡K(α),where S*(α) and
K(α)(0≤α<1) denote the subclasses of functions in 𝒜 which are, respectively, starlike of order α and convex of order α in U.
Also we have the class(2.3)S(m,λ,0;A,B)≡Sλm(A,B)studied by Patel [13].Letϕ(z) be analytic in U and ϕ(0)=1.
We introduce the following definition.Definition 2.3.
A functionf∈𝒜(n) is said to be in the class 𝒜(m,λ,l,n;ϕ) if it satisfies the following
subordination:(2.4)I(m+1,λ,l)f(z)I(m,λ,l)f(z)≺ϕ(z),(z∈U).Remark 2.4.
We note that the classes𝒜(m,1,l,n;ϕ) were investigated recently by Sivaprasad Kumar et al. [8].Theorem 2.5.
Let−1≤B<A≤1, l≥0, λ>0,
and(2.5)(1−B)(1−λ+l)+λ(1−A)>0.
(i) Then(2.6)S(m+1,λ,l;A,B)⊂S(m,λ,l;A,B).
Further,  for f∈S(m+1,λ,l;A,B),
the following
hold:
(2.7)l+1λ⋅I(m+1,λ,l)f(z)I(m,λ,l)f(z)−1−λ+ll+1≺q(z)≺1+Az1+Bz,where(2.8)q(z)={1(1/n)∫01s(l+1)/λn−1((1+Bzs)/(1+Bz))(1/n)(A/B−1)ds−1−λ+lλifB≠0,1(1/n)∫01s(l+1)/λn−1⋅exp⁡(Az(s−1)/n)ds−1−λ+lλifB=0,and q is the best dominant of (2.7).
(ii) Furthermore, in addition to (2.5), one consider
the inequality(2.9)A≤−B[l+1+λ(n−1)]λ,where −1≤B<0,
then(2.10)S(m+1,λ,l;A,B)⊂S(m,λ,l;1−2η,−1),where(2.11)η={[F21(1,1n(1−AB),l+1λn+1,BB−1)]−1−(1−λ+l)}/λ.
The result is the best possible.Proof.
Setting(2.12)χ(z):=z(I(m,λ,l)f(z))′I(m,λ,l)f(z)(z∈U),we note that χ(z) is analytic in U and(2.13)χ(z)=1+anzn+an+1zn+1+⋯.
Using the identity(2.14)(l+1)I(m+1,λ,l)f(z)=(1−λ+l)I(m,λ,l)f(z)+λz(I(m,λ,l)f(z))′in definition of χ(z) and carrying out logarithmic differentiation
in the resulting equation, one obtains(2.15)z(I(m+1,λ,l)f(z))′I(m+1,λ,l)f(z)=χ(z)+zχ′(z)χ(z)+(1−λ+l)/λ.
Since f∈S(m+1,λ,l;A,B),
we get(2.16)χ(z)+zχ′(z)χ(z)+(1−λ+l)/λ≺1+Az1+Bz.
By applying Lemma 1.4, we obtain that(2.17)χ(z)≺1+Az1+Bz.
Hence we have shown the inclusion (2.6). Also, making
use of Lemma 1.7 with β=1 and γ=(1−λ+l)/λ, q is the best dominant of (2.7) and q is defined by (2.8). This proves part (i) of
Theorem 2.5.
To establish (2.10), we need to show
that(2.18)inf⁡|z|<1Re{q(z)}=q(−1).The proof of the assertion
(2.18) will be deduced on the same lines as in [14] making use of
Lemma 1.5. If we set a=(1/n)(1−A/B),b=(l+1)/λn,c=(l+1)/λn+1, B<0,
then by using (1.13), (1.14), and
(1.15), we find from (2.8) that(2.19)q(z)=1(1/n)Q(z)−1−λ+lλ,where(2.20)Q(z)=(1+Bz)a∫01sb−1(1+Bsz)−ads=Γ(b)Γ(c)⋅2F1(1,a;c;BzBz+1).By using (1.13), the above
equality yields(2.21)Q(z)=∫01g(s,z)dμ(s),where(2.22)g(s,z)=1+Bz1+(1−s)Bz,dμ(s)=Γ(c)Γ(a)Γ(c−a)sa−1(1−s)c−a−1dsis a positive measure on the
closed interval [0,1].
For−1≤B<1,
we note that Reg(s,z)>0, g(s,−r) is real for 0≤r<1 and s∈[0,1] and(2.23)Re{1g(s,z)}≥1−(1−s)Br1−Br=1g(s,−r),|z|≤r<1.Therefore, by using Lemma 1.5,
one obtains(2.24)Re{1Q(z)}≥1Q(−r),|z|≤r<1which, upon letting r→1−1,
yields(2.25)Re{1Q(z)}≥1Q(−1).Now, the assertion (2.18)
follows by using Lemma 1.5. The result is the best possible and q is the best dominant of (2.7). This completes
the proof of Theorem 2.5.Takingm=0, n=1, λ=1, l=0, A=1−2α, B=−1 in Theorem 2.5, we get the following result
due to MacGregor [15].Corollary 2.6.
For0≤α<1,
one obtains(2.26)K(α)⊂S*(η1),where(2.27)η1=[2F1(1,2(1−α),2,12)]−1={1−2α22(1−α)(1−22α−1),α≠12,12ln⁡2,α=12.
The result is the best possible.Theorem 2.7.
Letψ(z) be univalent in U with ψ(0)=1, Reψ(z)>0,
and let
zψ′(z)/ψ(z) be starlike in U.
Let ϕ(z) be defined by(2.28)ϕ(z):=λl+1(l+1λψ(z)+zψ′(z)ψ(z)).Then(2.29)A(m+1,λ,l,n;ϕ)⊂A(m,λ,l,n;ψ).Proof.
Setting(2.30)q(z)=I(m+1,λ,l)f(z)I(m,λ,l)f(z),we note that q(z) is analytic in U.
By a simple computation, we observe from (2.30) that(2.31)zq′(z)q(z)=z(I(m+1,λ,l)f(z))′I(m+1,λ,l)f(z)−z(I(m,λ,l)f(z))′I(m,λ,l)f(z).
Making use of the identity (2.14), one obtains from
(2.31)(2.32)I(m+2,λ,l)f(z)I(m+1,λ,l)f(z)=λl+1(l+1λq(z)+zq′(z)q(z)).
By the hypothesis of Theorem 2.7 that f belongs to the class 𝒜(m+1,λ,l,n;ϕ) and in view of (2.32), we have(2.33)λl+1(l+1λq(z)+zq′(z)q(z))≺λl+1(l+1λψ(z)+zψ′(z)ψ(z)).
If we let(2.34)Q(z):=zψ′(z)φ(ψ(z)),where(2.35)φ(ψ(z))=λl+1⋅1ψ(z),h(z):=ψ(z)+Q(z)and since Q(z) is starlike, our theorem is an immediate
consequence of Lemma 1.6.Theorem 2.8.
Letψ be univalent in U, ψ(0)=1 and let
γ be a complex number. Suppose that
(1)
Re[λ(γ+1)−(l+1)+(l+1)ψ(z)]>0
and(2)
zψ′(z)/(λ(γ+1)−(l+1)+(l+1)ψ(z)) is starlike in U.
Let the functionF(z) be defined by(2.36)F(z):=γ+1zγ∫0ztγ−1f(t)dtand the function(2.37)h(z):=ψ(z)+λzψ′(z)λ(γ+1)−(l+1)+(l+1)ψ(z),then f∈𝒜(m,λ,l,n;h) implies F∈𝒜(m,λ,l,n;ψ).Proof.
From the definition ofF(z) and(2.38)(γ+1)I(m,λ,l)f(z)=l+1λI(m+1,λ,l)F(z)+(γ−1−λ+lλ)I(m,λ,l)F(z),if we let(2.39)q(z):=I(m+1,λ,l)F(z)I(m,λ,l)F(z),then we note that q(z) is analytic in U.
Using (2.38) and (2.39), one obtains(2.40)(γ+1)I(m,λ,l)f(z)I(m,λ,l)F(z)=γ−1−λ+lλ+l+1λq(z).
Differentiating this equality, we
obtain(2.41)I(m+1,λ,l)f(z)I(m,λ,l)f(z)=q(z)+λzq′(z)λ(γ+1)−(l+1)+(l+1)q(z).
For f∈𝒜(m,λ,l,n;h),
we have from (2.41)(2.42)q(z)+λzq′(z)λ(γ+1)−(l+1)+(l+1)q(z)≺ψ(z)+λzψ′(z)λ(γ+1)−(l+1)+(l+1)ψ(z).
If we let(2.43)Q(z):=zψ′(z)φ(ψ(z)),where(2.44)φ(ψ(z))=λλ(γ+1)−(l+1)+(l+1)ψ(z),h(z):=v(ψ(z))+Q(z),v(ψ(z)):=ψ(z)and since Q(z) is starlike in U,
our theorem is an immediate consequence of Lemma 1.6.Theorem 2.9.
Letf(z)∈𝒜(n).
Then f belongs to the class 𝒜(m,λ,l,n;χ) if and only if F(z) defined by(2.45)F(z):=l+1z(1−λ+l)/λ∫0zt(1−λ+l)/λ−1f(t)dtbelongs to the class 𝒜(m+1,λ,l,n;χ).Proof.
From the definition ofF(z),
we have(2.46)1−λ+lλF(z)+zF′(z)=(l+1)f(z).
By convoluting (2.46) with the function(2.47)u(m,λ,l,n;z):=z+∑k=n+1∞[1+λ(k−1)+ll+1]mzkand using a convolution property(2.48)z(f*g)′(z)=f(z)*zg′(z),one obtains(2.49)(l+1)I(m,λ,l)f(z)=1−λ+lλI(m,λ,l)F(z)+z(u(m,λ,l,n;z)*F(z))′,that is,(2.50)(l+1)I(m,λ,l)f(z)=1−λ+lλI(m,λ,l)F(z)+z(I(m,λ,l)F(z))′.
By using identity (2.14), we get(2.51)I(m,λ,l)f(z)=1λI(m+1,λ,l)F(z).
Also, we obtain(2.52)(l+1)I(m+1,λ,l)f(z)=(1−λ+l)I(m,λ,l)f(z)+λz(I(m,λ,l)f(z))′=1−λ+lλI(m+1,λ,l)F(z)+z(I(m+1,λ,l)F(z))′=l+1λI(m+2,λ,l)F(z).
From (2.51) and
(2.52), we get(2.53)I(m+2,λ,l)F(z)I(m+1,λ,l)F(z)=I(m+1,λ,l)f(z)I(m,λ,l)f(z).
By the hypothesis of Theorem 2.9 that(2.54)I(m+1,λ,l)f(z)I(m,λ,l)f(z)≺χ(z)and using (2.53), the desired
result follows at once.



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