Citation: Velislava N Lyubenova, Maya N Ignatova. On-line estimation of physiological states for monitoring and control of bioprocesses[J]. AIMS Bioengineering, 2017, 4(1): 93-112. doi: 10.3934/bioeng.2017.1.93
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Let H(U) be the class of analytic functions in the open unit disc U={z∈C:|z|<1} and let H[a,υ] be the subclass of H(U) including form-specific functions
f(z)=a+aυzυ+aυ+1zυ+1+…(a∈C), |
we denote by H=H[1,1].
Also, A(p) should denote the class of multivalent analytic functions in U, with the power series expansion of the type:
f(z)=zp+∞∑υ=p+1aυzυ(p∈N={1,2,3,..}). | (1.1) |
Upon differentiating j-times for each one of the (1.1) we obtain:
f(j)(z)=δ(p,j)zp−j+∞∑υ=p+1δ(υ,j)aυzυ−jz∈U,δ(p,j)=p!(p−j)! (p∈N, j∈N0=N∪{0}, p≥j). | (1.2) |
Numerous mathematicians, for instance, have looked at higher order derivatives of multivalent functions (see [1,3,6,9,16,27,28,31]).
For f,ℏ∈H, the function f is subordinate to ℏ or the function ℏ is said to be superordinate to f in U and we write f(z)≺ℏ(z), if there exists a Schwarz function ω in U with ω(0)=0 and |ω(z)|<1, such that f(z)=ℏ(ω(z)), z∈U. If ℏ is univalent in U, then f(z)≺ℏ(z) iff f(0)=ℏ(0) and f(U)⊂ℏ(U). (see [7,21]).
In the concepts and common uses of fractional calculus (see, for example, [14,15] see also [2]; the Riemann-Liouville fractional integral operator of order α∈C (ℜ(α)>0) is one of the most widely used operators (see [29]) given by:
(Iα0+f)(x)=1Γ(α)∫x0(x−μ)α−1f(μ)dμ(x>0;ℜ(α)>0) | (1.3) |
applying the well-known (Euler's) Gamma function Γ(α). The Erd élyi-Kober fractional integral operator of order α∈C(ℜ(α)>0) is an interesting alternative to the Riemann-Liouville operator Iα0+, defined by:
(Iα0+;σ,ηf)(x)=σx−σ(α+η)Γ(α)∫x0μσ(η+1)−1(xσ−μσ)α−1f(μ)dμ | (1.4) |
(x>0;ℜ(α)>0), |
which corresponds essentially to (1.3) when σ−1=η=0, since
(Iα0+;1,0f)(x)=x−α(Iα0+f)(x)(x>0;ℜ(α)>0). |
Mainly motivated by the special case of the definition (1.4) when x=σ=1, η=ν−1 and α=ρ−ν, here, we take a look at the integral operator ℑp(ν,ρ,μ) with f∈A(p) by (see [11])
ℑp(ν,ρ;ℓ)f(z)=Γ(ρ+ℓp)Γ(ν+ℓp)Γ(ρ−ν)∫10μν−1(1−μ)ρ−ν−1f(zμℓ)dμ |
(ℓ>0;ν,ρ∈R;ρ>ν>−ℓp;p∈N). |
Evaluating (Euler's) Gamma function by using the Eulerian Beta-function integral as following:
B(α,β):={∫10μα−1(1−μ)β−1dμ(min{ℜ(α),ℜ(β)}>0)Γ(α)Γ(β)Γ(α+β)(α,β∈C∖Z′0), |
we readily find that
ℑp(ν,ρ;ℓ)f(z)={zp+Γ(ρ+pℓ)Γ(ν+pℓ)∞∑υ=p+1Γ(ν+υℓ)Γ(ρ+υℓ)aυzυ(ρ>ν)f(z)(ρ=ν). | (1.5) |
It is readily to obtain from (1.5) that
z(ℑp(ν,ρ;ℓ)f(z))′=(νℓ+p)(ℑp(ν+1,ρ;ℓ)f(z))−νℓ(ℑp(ν,ρ;ℓ)f(z)). | (1.6) |
The integral operator ℑp(ν,ρ;ℓ)f(z) should be noted as a generalization of several other integral operators previously discussed for example,
(ⅰ) If we set p=1, we get ˜I(ν,ρ;ℓ)f(z) defined by Ŕaina and Sharma ([22] with m=0);
(ⅱ) If we set ν=β,ρ=β+1 and ℓ=1, we obtain ℑβpf(z)(β>−p) it was presented by Saitoh et al.[24];
(ⅲ) If we set ν=β,ρ=α+β−δ+1, ℓ=1, we obtain ℜα,δβ,pf(z)(δ>0; α≥δ−1; β>−p) it was presented by Aouf et al. [4];
(ⅳ) If we put ν=β,ρ=α+β, ℓ=1, we get Qαβ,pf(z)(α≥0;β>−p) it was investigated by Liu and Owa [18];
(ⅴ) If we put p=1, ν=β,ρ=α+β, ℓ=1, we obtain ℜαβf(z)(α≥0;β>−1) it was introduced by Jung et al. [13];
(ⅵ) If we put p=1, ν=α−1, ρ=β−1, ℓ=1, we obtain L(α,β)f(z)(α,β∈C∖Z0,Z0={0,−1,−2,...}) which was defined by Carlson and Shaffer [8];
(ⅶ) If we put p=1, ν=ν−1, ρ=j, ℓ=1 we obtain Iν,jf(z)(ν>0;j≥−1) it was investigated by Choi et al. [10];
(ⅷ) If we put p=1, ν=α,ρ=0, ℓ=1, we obtain Dαf(z)(α>−1) which was defined by Ruscheweyh [23];
(ⅸ) If we put p=1, ν=1, ρ=m, ℓ=1, we obtain Imf(z)(m∈N0) which was introduced by Noor [21];
(ⅹ) If we set p=1, ν=β,ρ=β+1, ℓ=1 we obtain ℑβf(z) which was studied by Bernadi [5];
(ⅹⅰ) If we set p=1, ν=1, ρ=2, ℓ=1 we get ℑf(z) which was defined by Libera [17].
We state various definition and lemmas which are essential to obtain our results.
Definition 1. ([20], Definition 2, p.817) We denote by Q the set of the functions f that are holomorphic and univalent on ¯U∖E(f), where
E(f)={ζ:ζ∈∂U and limz→ζf(z)=∞}, |
and satisfy f′(ζ)≠0 for ζ∈∂U∖E(f).
Lemma 1. ([12]; see also ([19], Theorem 3.1.6, p.71)) Assume that h(z) is convex (univalent) function in U with h(0)=1, and let φ(z)∈H, is analytic in U. If
φ(z)+1γzφ′(z)≺h(z)(z∈U), |
where γ≠0 and Re(γ)≥0. Then
φ(z)≺Ψ(z)=γzγz∫0tγ−1h(t)dt≺h(z)(z∈U), |
and Ψ(z) is the best dominant.
Lemma 2. ([26]; Lemma 2.2, p.3) Suppose that q is convex function in U and let ψ∈C with ϰ∈C∗=C∖{0} with
Re(1+zq′′(z)q′(z))>max{0;−Reψϰ},z∈U. |
If λ(z) is analytic in U, and
ψλ(z)+ϰzλ′(z)≺ψq(z)+ϰzq′(z), |
therefore λ(z)≺q(z), and q is the best dominant.
Lemma 3. ([20]; Theorem 8, p.822) Assume that q is convex univalent in U and suppose δ∈C, with Re(δ)>0. If λ∈H[q(0),1]∩Q and λ(z)+δzλ′(z) is univalent in U, then
q(z)+δzq′(z)≺λ(z)+δzλ′(z), |
implies
q(z)≺λ(z) (z∈U) |
and q is the best subordinant.
For a,ϱ,c and c(c∉Z−0) real or complex number the Gaussian hypergeometric function is given by
2F1(a,ϱ;c;z)=1+aϱc.z1!+a(a+1)ϱ(ϱ+1)c(c+1).z22!+.... |
The previous series totally converges for z∈U to a function analytical in U (see, for details, ([30], Chapter 14)) see also [19].
Lemma 4. For a,ϱ and c (c∉Z−0), real or complex parameters,
1∫0tϱ−1(1−t)c−ϱ−1(1−zt)−xdt=Γ(ϱ)Γ(c−a)Γ(c)2F1(a,ϱ;c;z)(Re(c)>Re(ϱ)>0); | (2.1) |
2F1(a,ϱ;c;z)=2F1(ϱ,a;c;z); | (2.2) |
2F1(a,ϱ;c;z)=(1−z)−a2F1(a,c−ϱ;c;zz−1); | (2.3) |
2F1(1,1;2;azaz+1)=(1+az)ln(1+az)az; | (2.4) |
2F1(1,1;3;azaz+1)=2(1+az)az(1−ln(1+az)az). | (2.5) |
Throughout the sequel, we assume unless otherwise indicated −1≤D<C≤1, δ>0, ℓ>0, ν,ρ∈R, ν>−ℓp, p∈N and (ρ−j)≥0. We shall now prove the subordination results stated below:
Theorem 1. Let 0≤j<p, 0<r≤1 and for f∈A(p) assume that
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U, | (3.1) |
whenever δ∈(0,+∞)∖N. Let define the function Φj by
Φj(z)=(1−α)((ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ+α(ℑp(ν+1,ρ;ℓ)f(z))(j)zp−j((ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ−1, |
such that the powers are all the principal ones, i.e., log1 = 0. Whether
Φj(z)≺[p!(p−j)!]δ(1+Cz1+Dz )r, | (3.2) |
then
((ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ≺[p!(p−j)!]δp(z), | (3.3) |
where
p(z)={(CD)r∑i≥0(−r)ii!(C−DC)i(1+Dz)−i 2F1(i,1;1+δ(ν+ℓp)αℓ;Dz1+Dz)(D≠0);2F1(−r,δ(ν+ℓp)αℓ;1+δ(ν+ℓp)αℓ;−Cz) (D=0), |
and [p!(p−j)!]δp(z) is the best dominant of (3.3). Moreover, there are
ℜ((ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ>[p!(p−j)!]δζ, z∈U, | (3.4) |
where ζ is given by:
ζ={(CD)r∑i≥0(−r)ii!(C−DC)i(1−D)−i 2F1(i,1;1+δ(ν+ℓp)αℓ;DD−1)(D≠0);2F1(−r,δ(ν+ℓp)αℓ;1+δ(ν+ℓp)αℓ;C) (D=0), |
then (3.4) is the best possible.
Proof. Let
ϕ(z)=((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ, (z∈U). | (3.5) |
It is observed that the function ϕ(z)∈H, which is analytic in U and ϕ(0)=1. Differentiating (3.5) with respect to z, applying the given equation, the hypothesis (3.2), and the knowing that
z(ℑp(ν,ρ;ℓ)f(z))(j+1)=(νℓ+p)(ℑp(ν+1,ρ;ℓ)f(z))(j)−(νℓ+j)(ℑp(ν,ρ;ℓ)f(z))(j) (0≤j<p), | (3.6) |
we get
ϕ(z)+zϕ′(z)δ(ν+ℓp)αℓ≺(1+Cz1+Dz )r=q(z) (z∈U). |
We can verify that the above equation q(z) is analytic and convex in U as following
Re(1+zq′′(z)q′(z))=−1+(1−r)ℜ(11+Cz)+(1+r)ℜ(11+Dz)>−1+1−r1+|C|+1+r1+|D|≥0 (z∈U). |
Using Lemma 1, there will be
ϕ(z)≺p(z)=δ(ν+ℓp)αℓz−δ(ν+ℓp)αℓz∫0tδ(ν+ℓp)αℓ−1(1+Ct1+Dt)rdt. |
In order to calculate the integral, we define the integrand in the type
tδ(ν+ℓp)αℓ−1(1+Ct1+Dt)r=tδ(ν+ℓp)αℓ−1(CD)r(1−C−DC+CDt)r, |
using Lemma 4 we obtain
p(z)=(CD)r∑i≥0(−r)ii!(C−DC)i(1+Dz)−i 2F1(i,1;1+δ(ν+ℓp)αℓ;Dz1+Dz)(D≠0). |
On the other hand if D=0 we have
p(z)=2F1(−r,δ(ν+ℓp)αℓ;1+δ(ν+ℓp)αℓ;−Cz), |
where the identities (2.1)–(2.3), were used after changing the variable, respectively. This proof the inequality (3.3).
Now, we'll verify it
inf{ℜp(z):|z|<1}=p(−1). | (3.7) |
Indeed, we have
ℜ(1+Cz1+Dz )r≥(1−Cσ1−Dσ)r (|z|<σ<1). |
Setting
ℏ(s,z)=(1+Csz1+Dsz)r (0≤s≤1; z∈U) |
and
dv(s)=δ(ν+ℓp)αℓsδ(ν+ℓp)αℓ−1ds |
where dv(s) is a positive measure on the closed interval [0, 1], we get that
p(z)=1∫0ℏ(s,z)dv(s), |
so that
ℜp(z)≥1∫0(1−Csσ1−Dsσ)rdv(s)=p(−σ) (|z|<σ<1). |
Now, taking σ→1− we get the result (3.7). The inequality (3.4) is the best possible since [p!(p−j)!]δp(z) is the best dominant of (3.3).
If we choose j=1 and α=δ=1 in Theorem 1, we get:
Corollary 1. Let 0<r≤1. If
(ℑp(ν+1,ρ;ℓ)f(z))′zp−1≺p(1+Cz1+Dz )r, |
then
ℜ((ℑp(ν,ρ;ℓ)f(z))′zp−1)>pζ1, z∈U, | (3.8) |
where ζ1 is given by:
ζ1={(CD)r∑i≥0(−r)ii!(C−DC)i(1−D)−i 2F1(i,1;1+(ν+ℓp)ℓ;DD−1)(D≠0);2F1(−r,(ν+ℓp)ℓ;1+(ν+ℓp)ℓ;C) (D=0), |
then (3.8) is the best possible.
If we choose ν=ρ=0 and ℓ=1 in Theorem 1, we get:
Corollary 2. Let 0≤j<p, 0<r≤1 and as f∈A(p) assume that
f(j)(z)zp−j≠0, z∈U, |
whenever δ∈(0,+∞)∖N. Let define the function Φj by
Φj(z)=[1−α(1−jp)](f(j)(z)zp−j)δ+α(zf(j+1)(z)pf(j)(z))(f(j)(z)zp−j)δ, | (3.9) |
such that the powers are all the principal ones, i.e., log1 = 0. If
Φj(z)≺[p!(p−j)!]δ(1+Cz1+Dz )r, |
then
(f(j)(z)zp−j)δ≺[p!(p−j)!]δp1(z), | (3.10) |
where
p1(z)={(CD)r∑i≥0(−r)ii!(C−DC)i(1+Dz)−i 2F1(i,1;1+δpα;Dz1+Dz)(D≠0);2F1(−r,δpα;1+δpα;−Cz) (D=0), |
and [p!(p−j)!]δp1(z) is the best dominant of (3.10). Morover, there are
ℜ(f(j)(z)zp−j)δ>[p!(p−j)!]δζ2, z∈U, | (3.11) |
where ζ2 is given by
ζ2={(CD)r∑i≥0(−r)ii!(C−DC)i(1−D)−i 2F1(i,1;1+δpα;DD−1)(D≠0);2F1(−r,δpα;1+δpα;C) (D=0), |
then (3.11) is the best possible.
If we put δ=1 and r=1 in Corollary 2, we get:
Corollary 3. Let 0≤j<p, and for f∈A(p) say it
f(j)(z)zp−j≠0, z∈U. |
Let define the function Φj by
Φj(z)=[(1−α(1−jp)]f(j)(z)zp−j+αf(j+1)(z)pzp−j−1. |
If
Φj(z)≺p!(p−j)!1+Cz1+Dz, |
then
f(j)(z)zp−j≺p!(p−j)!p2(z), | (3.12) |
where
p2(z)={CD+(1−CD)(1+Dz)−1 2F1(1,1;1+pα;Dz1+Dz)(D≠0);1+pp+αCz, (D=0), |
and p!(p−j)!p2(z) is the best dominant of (3.12). Morover there will be
ℜ(f(j)(z)zp−j)>p!(p−j)!ζ3, z∈U, | (3.13) |
where ζ3 is given by:
ζ3={CD+(1−CD)(1−D)−1 2F1(1,1;1+pα;DD−1)(D≠0);1−pp+αC, (D=0), |
then (3.13) is the best possible.
For C=1,D=−1 and j=1 Corollary 3, leads to the next example:
Example 1. (i) For f∈A(p) suppose that
f′(z)zp−1≠0, z∈U. |
Let define the function Φj by
Φj(z)=[1−(α−αp)]f′(z)zp−1+αf′′(z)pzp−2≺p1+z1−z, |
then
f′(z)zp−1≺p1+z1−z, | (3.14) |
and
ℜ(f′(z)zp−1)>pζ4, z∈U, | (3.15) |
where ζ4 is given by:
ζ4=−1+ 2F1(1,1;p+αα;12), |
then (3.15) is the best possible.
(ii) For p=α=1, (i) leads to:
For f∈A suppose that
f′(z)≠0, z∈U. |
Let define the function Φj by
Φj(z)=f′(z)+zf′′(z)≺1+z1−z, |
then
ℜ(f′(z))>−1+2ln2, z∈U. |
So the estimate is best possible.
Theorem 2. Let 0≤j<p, 0<r≤1 as for f∈A(p). Assume that Fα is defined by
Fα(z)=α(νℓ+p)(ℑp(ν+1,ρ;ℓ)f(z))+(1−α−α(νℓ))(ℑp(ν,ρ;ℓ)f(z)). | (3.16) |
If
F(j)α(z)zp−j≺(1−α+αp)p!(p−j)!(1+Cz1+Dz )r, | (3.17) |
then
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≺p!(p−j)!p(z), | (3.18) |
where
p(z)={(CD)r∑i≥0(−r)ii!(C−DC)i(1+Dz)−i 2F1(i,1;1+(1−α+αp)α;Dz1+Dz)(D≠0);2F1(−r,(1−α+αp)α;1+(1−α+αp)α;−Cz) (D=0), |
and p!(p−j)!p(z) is the best dominant of (3.18). Moreover, there will be
ℜ((ℑp(ν,ρ;ℓ)f(z))(j)zp−j)>p!(p−j)!η, z∈U, | (3.19) |
where η is given by:
η={(CD)r∑i≥0(−r)ii!(C−DC)i(1+D)−i 2F1(i,1;1+(1−α+αp)α;DD−1)(D≠0);2F1(−r,(1−α+αp)α;1+(1−α+αp)α;C) (D=0), |
then (3.19) is the best possible.
Proof. By using the definition (3.16) and the inequality (3.6), we have
F(j)α(z)=αz(ℑp(ν,ρ;ℓ)f(z))(j+1)+(1−α+αj)(ℑp(ν,ρ;ℓ)f(z))(j), | (3.20) |
for 0≤j<p. Putting
ϕ(z)=(p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j, (z∈U), | (3.21) |
we have that ϕ∈H. Differentiating (3.21), and using (3.17), (3.20), we get
ϕ(z)+zϕ′(z)(1−α+αp)α≺(1+Cz1+Dz )r (z∈U). |
Following the techniques of Theorem 1, we can obtain the remaining part of the proof.
If we choose j=1 and r=1 in Theorem 2, we get:
Corollary 4. For f∈A(p) let the function Fα define by 3.16. If
F′α(z)zp−1≺p(1−α+αp)1+Cz1+Dz , |
then
ℜ((ℑp(ν,ρ;ℓ)f(z))′zp−1)>pη1, z∈U, | (3.22) |
where η1 is given by:
η1={CD+(1−CD)(1−D)−1 2F1(1,1;1+1−α+αpα;DD−1)(D≠0);1−1−α+αp1+αpC (D=0), |
then (3.22) is the best possible.
Example 2. If we choose p=C=α=1 and D=−1 in Corollary 4, we obtain:
For
F(z)=(νℓ+1)(ℑ(ν+1,ρ;ℓ)f(z))−(νℓ)(ℑ(ν,ρ;ℓ)f(z)). |
If
F′(z)≺1+z1−z, |
then
ℜ((ℑ(ν,ρ;ℓ)f(z))′)>−1+2ln2, z∈U, |
the result is the best possible.
Theorem 3. Let 0≤j<p, 0<r≤1 as for θ>−p assume that Jp,θ:A(p)→A(p) defined by
Jp,θ(f)(z)=p+θzθz∫0tθ−1f(t)dt, z∈U. | (3.23) |
If
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≺p!(p−j)!(1+Cz1+Dz )r, | (3.24) |
then
(ℑp(ν,ρ;ℓ)Jp,θ(f)(z))(j)zp−j≺p!(p−j)!p(z), | (3.25) |
where
p(z)={(CD)r∑i≥0(−r)ii!(C−DC)i(1+Dz)−i 2F1(i,1;1+θ+p;Dz1+Dz)(D≠0);2F1(−r,θ+p;1+θ+p;Cz) (D=0), |
and p!(p−j)!p(z) is the best dominant of (3.25). Moreover, there will be
ℜ((ℑp(ν,ρ;ℓ)Jp,θ(f)(z))(j)zp−j)>p!(p−j)!β, z∈U, | (3.26) |
where β is given by:
β={(CD)r∑i≥0(−r)ii!(C−DC)i(1+D)−i 2F1(i,1;1+θ+p;DD−1)(D≠0);2F1(−r,θ+p;1+θ+p;−C) (D=0), |
then (3.26) is the best possible.
Proof. Suppose
ϕ(z)=(p−j)!p!(ℑp(ν,ρ;ℓ)Jp,θ(f)(z))(j)zp−j, (z∈U), |
we have that ϕ∈H. Differentiating the above definition, by using (3.24) and
z(ℑp(ν,ρ;ℓ)Jp,θ(f)(z))(j+1)=(θ+p)(ℑp(ν,ρ;ℓ)f(z))(j)−(θ+j)(ℑp(ν,ρ;ℓ)Jp,θ(f)(z))(j) (0≤j<p), |
we get
ϕ(z)+zϕ′(z)θ+p≺(1+Cz1+Dz )r. |
Now, we obtain (3.25) and the inequality (3.26) follow by using the same techniques in Theorem 1.
If we set j=1 and r=1 in Theorem 3, we get:
Corollary 5. For θ>−p, let the operator Jp,θ:A(p)→A(p) defined by (3.25). If
(ℑp(ν,ρ;ℓ)f(z))′zp−1≺p1+Cz1+Dz , |
then
ℜ((ℑp(ν,ρ;ℓ)Jp,θ(f)(z))′zp−1)>pβ1, z∈U, | (3.27) |
where β1 is given by:
β1={CD+(1−CD)(1−D)−1 2F1(1,1;1+θ+p;DD−1)(D≠0);1−θ+p1+θ+pC (D=0), |
then (3.27) is the best possible.
Example 3. If we choose p=C=θ=1 and D=−1 in Corollary 5, we get:
If
(ℑ(ν,ρ;ℓ)f(z))′≺1+z1−z, |
then
ℜ((ℑ(ν,ρ;ℓ)J1,1(f)(z))′)>−1+4(1−ln2), |
the result is the best possible.
Theorem 4. Let q is univalent function in U, such that q satisfies
Re(1+zq′′(z)q′(z))>max{0;−δ(ν+ℓp)αℓ}, z∈U. | (3.28) |
Let 0≤j<p, 0<r≤1 and for f∈A(p) assume that
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U, |
whenever δ∈(0,+∞)∖N. Let the function Φj defined by (3.1), and assume that it satisfies:
[(p−j)!p!]δΦj(z)≺q(z)+αℓδ(ν+ℓp)zq′(z). | (3.29) |
Then,
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ≺q(z), | (3.30) |
and q(z) is the best dominant of (3.30).
Proof. Let ϕ(z) is defined by (3.5), from Theorem 1 we get
[(p−j)!p!]δΦj(z)=ϕ(z)+αℓδ(ν+ℓp)zϕ′(z). | (3.31) |
Combining (3.29) and (3.31) we find that
ϕ(z)+αℓδ(ν+ℓp)zϕ′(z)≺q(z)+αℓδ(ν+ℓp)zq′(z). | (3.32) |
The proof of Theorem 4 follows by using Lemma 2 and (3.32).
Taking q(z)=(1+Cz1+Dz)r in Theorem 4, we obtain:
Corollary 6. Suppose that
Re(1−Dz1+Dz+(r−1)(C−D)z(1+Dz)(1+Cz))>max{0;−δ(ν+ℓp)αℓ}, z∈U. |
Let 0≤j<p, 0<r≤1 and for f∈A(p) satisfies
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U, |
whenever δ∈(0,+∞)∖N. Let the function Φj defined by (3.1), satisfies:
[(p−j)!p!]δΦj(z)≺(1+Cz1+Dz )r+αℓδ(ν+ℓp)(1+Cz1+Dz )rr(C−D)z(1+Dz)(1+Cz). |
Then,
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ≺(1+Cz1+Dz )r, | (3.33) |
so (1+Cz1+Dz)r is the best dominant of (3.33).
Taking q(z)=1+Cz1+Dz in Theorem 4, we get:
Corollary 7. Suppose that
Re(1−Dz1+Dz)>max{0;−δ(ν+ℓp)αℓ}, z∈U. |
Let 0≤j<p, 0<r≤1 and for f∈A(p) satisfies
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U, |
whenever δ∈(0,+∞)∖N. Let the function Φj defined by (3.1), satisfies:
[(p−j)!p!]δΦj(z)≺1+Cz1+Dz +αℓδ(ν+ℓp)(C−D)z(1+Dz)2. |
Then,
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ≺1+Cz1+Dz , | (3.34) |
so 1+Cz1+Dz is the best dominant of (3.34).
If we put ν=ρ=0 and ℓ=1 in Theorem 4, we get:
Corollary 8. Let q is univalent function in U, such that q satisfies
Re(1+zq′′(z)q′(z))>max{0;−δpα}, z∈U. |
For f∈A(p) satisfies
f(j)(z)zp−j≠0, z∈U. |
Let the function Φj defined by (3.9), satisfies:
[(p−j)!p!]δΦj(z)≺q(z)+αδpzq′(z). | (3.35) |
Then,
((p−j)!p!f(j)(z)zp−j)δ≺q(z), | (3.36) |
so q(z) is the best dominant of (3.36).
Taking C=1 and D=−1 in Corollaries 6 and 7 we get:
Example 4. (i) For f∈A(p) assume that
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U. |
Let the function Φj defined by (3.1), and assume that it satisfies:
[(p−j)!p!]δΦj(z)≺(1+z1−z)r+αℓδ(ν+ℓp)(1+z1−z)r2rz1−z2. |
Then,
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ≺(1+z1−z)r, | (3.37) |
so (1+z1−z)r is the best dominant of (3.37).
(ii) For f∈A(p) say it
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U. |
Let the function Φj defined by (3.1), and assume that it satisfies:
[(p−j)!p!]δΦj(z)≺1+z1−z+αℓδ(ν+ℓp)2z1−z2. |
Then,
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ≺1+z1−z, | (3.38) |
so 1+z1−z is the best dominant of (3.38).
If we put p=C=α=δ=1, D=−1 and j=0 in Corollary 8 we get:
Example 5. For f∈A suppose that
f(z)z≠0, z∈U, |
and
f′(z)≺(1+z1−z)r+(1+z1−z)r2rz1−z2. |
Then,
f(z)z≺(1+z1−z)r, | (3.39) |
and (1+z1−z)r is the best dominant of (3.39).
Remark 1. For ν=ρ=0, ℓ=p=r=1 and j=0 in Theorem 4, we get the results investigated by Shanmugam et al. ([25], Theorem 3.1).
Theorem 5. Let 0≤j<p, and for f∈A(p) assume that
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U, |
whenever δ∈(0,+∞)∖N. Suppose that
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ∈H∩Q |
such that [(p−j)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). If q is convex (univalent) function in U, and
q(z)+αℓδ(ν+ℓp)zq′(z)≺[(p−j)!p!]δΦj(z), |
then
q(z)≺((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ, | (3.40) |
so q(z) is the best subordinate of (3.40).
Proof. Let ϕ is defined by (3.5), from (3.31) we get
q(z)+αℓδ(ν+ℓp)zq′(z)≺[(p−j)!p!]δΦj(z)=ϕ(z)+αℓδ(ν+ℓp)zϕ′(z). |
The proof of Theorem 5 followes by an application of Lemma 3.
Taking q(z)=(1+Cz1+Dz)r in Theorem 5, we get:
Corollary 9. Let 0≤j<p, 0<r≤1 and for f∈A(p) assume that
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U. |
Suppose that
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ∈H∩Q |
such that [(p−j)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). If
(1+Cz1+Dz )r+αℓδ(ν+ℓp)(1+Cz1+Dz )rr(C−D)z(1+Dz)(1+Cz)≺[(p−j)!p!]δΦj(z), |
then
(1+Cz1+Dz )r≺((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ, | (3.41) |
so (1+Cz1+Dz)r is the best dominant of (3.41).
Taking q(z)=1+Cz1+Dz and r=1 in Theorem 5, we get:
Corollary 10. Let 0≤j<p, and for f∈A(p) assume that
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U, |
whenever δ∈(0,+∞)∖N. Assume that
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ∈H∩Q |
such that [(p−j)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). If
1+Cz1+Dz +αℓδ(ν+ℓp)(C−D)z(1+Dz)2≺[(p−j)!p!]δΦj(z), |
then
1+Cz1+Dz ≺((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ, | (3.42) |
so 1+Cz1+Dz is the best dominant of (3.42).
Combining results of Theorems 4 and 5, we have
Theorem 6. Let 0≤j<p, and for f∈A(p) assume that
(ℑp(ν,ρ;ℓ)f(z))(j)zp−j≠0, z∈U. |
Suppose that
((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ∈H[q(0),1]∩Q |
such that [(p−j)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). Let q1 is convex (univalent) function in U, and assume that q2 is convex in U, that q2 satisfies (3.28). If
q1(z)+αℓδ(ν+ℓp)zq′1(z)≺[(p−j)!p!]δΦj(z)≺q2(z)+αℓδ(ν+ℓp)zq′2(z), |
then
q1(z)≺((p−j)!p!(ℑp(ν,ρ;ℓ)f(z))(j)zp−j)δ≺q2(z) |
and q1(z) and q2(z) are respectively the best subordinate and best dominant of the above subordination.
We used the application of higher order derivatives to obtained a number of interesting results concerning differential subordination and superordination relations for the operator ℑp(ν,ρ;ℓ)f(z) of multivalent functions analytic in U, the differential subordination outcomes are followed by some special cases and counters examples. Differential sandwich-type results have been obtained. Our results we obtained are new and could help the mathematicians in the field of Geometric Function Theory to solve other special results in this field.
This research has been funded by Deputy for Research & innovation, Ministry of Education through initiative of institutional funding at university of Ha'il, Saudi Arabia through project number IFP-22155.
The authors declare no conflict of interest.
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