Citation: Kottakkaran Sooppy Nisar, Gauhar Rahman, Aftab Khan, Asifa Tassaddiq, Moheb Saad Abouzaid. Certain generalized fractional integral inequalities[J]. AIMS Mathematics, 2020, 5(2): 1588-1602. doi: 10.3934/math.2020108
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Let A denote the class of functions f which are analytic in the open unit disk Δ={z∈C:|z|<1}, normalized by the conditions f(0)=f′(0)−1=0. So each f∈A has series representation of the form
f(z)=z+∞∑n=2anzn. | (1.1) |
For two analytic functions f and g, f is said to be subordinated to g (written as f≺g) if there exists an analytic function ω with ω(0)=0 and |ω(z)|<1 for z∈Δ such that f(z)=(g∘ω)(z).
A function f∈A is said to be in the class S if f is univalent in Δ. A function f∈S is in class C of normalized convex functions if f(Δ) is a convex domain. For 0≤α≤1, Mocanu [23] introduced the class Mα of functions f∈A such that f(z)f′(z)z≠0 for all z∈Δ and
ℜ((1−α)zf′(z)f(z)+α(zf′(z))′f′(z))>0(z∈Δ). | (1.2) |
Geometrically, f∈Mα maps the circle centred at origin onto α-convex arcs which leads to the condition (1.2). The class Mα was studied extensively by several researchers, see [1,10,11,12,24,25,26,27] and the references cited therein.
A function f∈S is uniformly starlike if f maps every circular arc Γ contained in Δ with center at ζ ∈Δ onto a starlike arc with respect to f(ζ). A function f∈C is uniformly convex if f maps every circular arc Γ contained in Δ with center ζ ∈Δ onto a convex arc. We denote the classes of uniformly starlike and uniformly convex functions by UST and UCV, respectively. For recent study on these function classes, one can refer to [7,9,13,19,20,31].
In 1999, Kanas and Wisniowska [15] introduced the class k-UCV (k≥0) of k-uniformly convex functions. A function f∈A is said to be in the class k-UCV if it satisfies the condition
ℜ(1+zf″(z)f′(z))>k|zf′(z)f′(z)|(z∈Δ). | (1.3) |
In recent years, many researchers investigated interesting properties of this class and its generalizations. For more details, see [2,3,4,14,15,16,17,18,30,32,35] and references cited therein.
In 2015, Sokół and Nunokawa [33] introduced the class MN, a function f∈MN if it satisfies the condition
ℜ(1+zf″(z)f′(z))>|zf′(z)f(z)−1|(z∈Δ). |
In [28], it is proved that if ℜ(f′)>0 in Δ, then f is univalent in Δ. In 1972, MacGregor [21] studied the class B of functions with bounded turning, a function f∈B if it satisfies the condition ℜ(f′)>0 for z∈Δ. A natural generalization of the class B is B(δ1) (0≤δ1<1), a function f∈B(δ1) if it satisfies the condition
ℜ(f′(z))>δ1(z∈Δ;0≤δ1<1), | (1.4) |
for details associated with the class B(δ1) (see [5,6,34]).
Motivated essentially by the above work, we now introduce the following class k-Q(α) of analytic functions.
Definition 1. Let k≥0 and 0≤α≤1. A function f∈A is said to be in the class k-Q(α) if it satisfies the condition
ℜ((zf′(z))′f′(z))>k|(1−α)f′(z)+α(zf′(z))′f′(z)−1|(z∈Δ). | (1.5) |
It is worth mentioning that, for special values of parameters, one can obtain a number of well-known function classes, some of them are listed below:
1. k-Q(1)=k-UCV;
2. 0-Q(α)=C.
In what follows, we give an example for the class k-Q(α).
Example 1. The function f(z)=z1−Az(A≠0) is in the class k-Q(α) with
k≤1−b2b√b(1+α)[b(1+α)+2]+4(b=|A|). | (1.6) |
The main purpose of this paper is to establish several interesting relationships between k-Q(α) and the class B(δ) of functions with bounded turning.
To prove our main results, we need the following lemmas.
Lemma 1. ([8]) Let h be analytic in Δ with h(0)=1, β>0 and 0≤γ1<1. If
h(z)+βzh′(z)h(z)≺1+(1−2γ1)z1−z, |
then
h(z)≺1+(1−2δ)z1−z, |
where
δ=(2γ1−β)+√(2γ1−β)2+8β4. | (2.1) |
Lemma 2. Let h be analytic in Δ and of the form
h(z)=1+∞∑n=mbnzn(bm≠0) |
with h(z)≠0 in Δ. If there exists a point z0(|z0|<1) such that |argh(z)|<πρ2(|z|<|z0|) and |argh(z0)|=πρ2 for some ρ>0, then z0h′(z0)h(z0)=iℓρ, where
ℓ:{ℓ≥n2(c+1c)(argh(z0)=πρ2),ℓ≤−n2(c+1c)(argh(z0)=−πρ2), |
and (h(z0))1/ρ=±ic(c>0).
This result is a generalization of the Nunokawa's lemma [29].
Lemma 3. ([37]) Let ε be a positive measure on [0,1]. Let ϝ be a complex-valued function defined on Δ×[0,1] such that ϝ(.,t) is analytic in Δ for each t∈[0,1] and ϝ(z,.) is ε-integrable on [0,1] for all z∈Δ. In addition, suppose that ℜ(ϝ(z,t))>0, ϝ(−r,t) is real and ℜ(1/ϝ(z,t))≥1/ϝ(−r,t) for |z|≤r<1 and t∈[0,1]. If ϝ(z)=∫10ϝ(z,t)dε(t), then ℜ(1/ϝ(z))≥1/ϝ(−r).
Lemma 4. ([22]) If −1≤D<C≤1, λ1>0 and ℜ(γ2)≥−λ1(1−C)/(1−D), then the differential equation
s(z)+zs′(z)λ1s(z)+γ2=1+Cz1+Dz(z∈Δ) |
has a univalent solution in Δ given by
s(z)={zλ1+γ2(1+Dz)λ1(C−D)/Dλ1∫z0tλ1+γ2−1(1+Dt)λ1(C−D)/Ddt−γ2λ1(D≠0),zλ1+γ2eλ1Czλ1∫z0tλ1+γ2−1eλ1Ctdt−γ2λ1(D=0). |
If r(z)=1+c1z+c2z2+⋯ satisfies the condition
r(z)+zr′(z)λ1r(z)+γ2≺1+Cz1+Dz(z∈Δ), |
then
r(z)≺s(z)≺1+Cz1+Dz, |
and s(z) is the best dominant.
Lemma 5. ([36,Chapter 14]) Let w, x and\ y≠0,−1,−2,… be complex numbers. Then, for ℜ(y)>ℜ(x)>0, one has
1. 2G1(w,x,y;z)=Γ(y)Γ(y−x)Γ(x)∫10sx−1(1−s)y−x−1(1−sz)−wds;
2. 2G1(w,x,y;z)= 2G1(x,w,y;z);
3. 2G1(w,x,y;z)=(1−z)−w2G1(w,y−x,y;zz−1).
Firstly, we derive the following result.
Theorem 1. Let 0≤α<1 and k≥11−α. If f∈k-Q(α), then f∈B(δ), where
δ=(2μ−λ)+√(2μ−λ)2+8λ4(λ=1+αkk(1−α);μ=k−αk−1k(1−α)). | (3.1) |
Proof. Let f′=ℏ, where ℏ is analytic in Δ with ℏ(0)=1. From inequality (1.5) which takes the form
ℜ(1+zℏ′(z)ℏ(z))>k|(1−α)ℏ(z)+α(1+zℏ′(z)ℏ(z))−1|=k|1−α−ℏ(z)+αℏ(z)−αzℏ′(z)ℏ(z)|, |
we find that
ℜ(ℏ(z)+1+αkk(1−α)zℏ(z)ℏ(z))>k−αk−1k(1−α), |
which can be rewritten as
ℜ(ℏ(z)+λzℏ(z)ℏ(z))>μ(λ=1+αkk(1−α);μ=k−αk−1k(1−α)). |
The above relationship can be written as the following Briot-Bouquet differential subordination
ℏ(z)+λzℏ′(z)ℏ(z)≺1+(1−2μ)z1−z. |
Thus, by Lemma 1, we obtain
ℏ≺1+(1−2δ)z1−z, | (3.2) |
where δ is given by (3.1). The relationship (3.2) implies that f∈B(δ). We thus complete the proof of Theorem 3.1.
Theorem 2. Let 0<α≤1, 0<β<1, c>0, k≥1, n≥m+1(m∈ N ), |ℓ|≥n2(c+1c) and
|αβℓ±(1−α)cβsinβπ2|≥1. | (3.3) |
If
f(z)=z+∞∑n=m+1anzn(am+1≠0) |
and f∈k-Q(α), then f∈B(β0), where
β0=min{β:β∈(0,1)} |
such that (3.3) holds.
Proof. By the assumption, we have
f′(z)=ℏ(z)=1+∞∑n=mcnzn(cm≠0). | (3.4) |
In view of (1.5) and (3.4), we get
ℜ(1+zℏ′(z)ℏ(z))>k|(1−α)ℏ(z)+α(1+zℏ′(z)ℏ(z))−1|. |
If there exists a point z0∈Δ such that
|argℏ(z)|<βπ2(|z|<|z0|;0<β<1) |
and
|argℏ(z0)|=βπ2(0<β<1), |
then from Lemma 2, we know that
z0ℏ′(z0)ℏ(z0)=iℓβ, |
where
(ℏ(z0))1/β=±ic(c>0) |
and
ℓ:{ℓ≥n2(c+1c)(argℏ(z0)=βπ2),ℓ≤−n2(c+1c)(argℏ(z0)=−βπ2). |
For the case
argℏ(z0)=βπ2, |
we get
ℜ(1+z0ℏ′(z0)ℏ(z0))=ℜ(1+iℓβ)=1. | (3.5) |
Moreover, we find from (3.3) that
k|(1−α)ℏ(z0)+α(1+z0ℏ′(z0)ℏ(z0))−1|=k|(1−α)(ℏ(z0)−1)+αz0ℏ′(z0)ℏ(z0)|=k|(1−α)[(±ic)β−1]+iαβℓ|=k√(1−α)2(cβcosβπ2−1)2+[αβℓ±(1−α)cβsinβπ2]2≥1. | (3.6) |
By virtue of (3.5) and (3.6), we have
ℜ(1+zℏ′(z0)ℏ(z0))≤k|(1−α)ℏ(z0)+α(1+z0ℏ(z0)ℏ(z0))−1|, |
which is a contradiction to the definition of k-Q(α). Since β0=min{β:β∈(0,1)} such that (3.3) holds, we can deduce that f∈B(β0).
By using the similar method as given above, we can prove the case
argℏ(z0)=−βπ2 |
is true. The proof of Theorem 2 is thus completed.
Theorem 3. If 0<β<1 and 0≤ν<1. If f∈k-Q(α), then
ℜ(f′)>[2G1(2β(1−ν),1;1β+1;12)]−1, |
or equivalently, k-Q(α)⊂B(ν0), where
ν0=[2G1(2β(1−μ),1;1β+1;12)]−1. |
Proof. For
w=2β(1−ν), x=1β, y=1β+1, |
we define
ϝ(z)=(1+Dz)w∫10tx−1(1+Dtz)−wdt=Γ(x)Γ(y) 2G1(1,w,y;zz−1). | (3.7) |
To prove k-Q(α)⊂B(ν0), it suffices to prove that
inf|z|<1{ℜ(q(z))}=q(−1), |
which need to show that
ℜ(1/ϝ(z))≥1/ϝ(−1). |
By Lemma 3 and (3.7), it follows that
ϝ(z)=∫10ϝ(z,t)dε(t), |
where
ϝ(z,t)=1−z1−(1−t)z(0≤t≤1), |
and
dε(t)=Γ(x)Γ(w)Γ(y−w)tw−1(1−t)y−w−1dt, |
which is a positive measure on [0,1].
It is clear that ℜ(ϝ(z,t))>0 and ϝ(−r,t) is real for |z|≤r<1 and t∈[0,1]. Also
ℜ(1ϝ(z,t))=ℜ(1−(1−t)z1−z)≥1+(1−t)r1+r=1ϝ(−r,t) |
for |z|≤r<1. Therefore, by Lemma 3, we get
ℜ(1/ϝ(z))≥1/ϝ(−r). |
If we let r→1−, it follows that
ℜ(1/ϝ(z))≥1/ϝ(−1). |
Thus, we deduce that k-Q(α)⊂B(ν0).
Theorem 4. Let 0≤α<1 and k≥11−α. If f∈k-Q(α), then
f′(z)≺s(z)=1g(z), |
where
g(z)=2G1(2λ,1,1λ+1;zz−1)(λ=1+αkk(1−α)). |
Proof. Suppose that f′=ℏ. From the proof of Theorem 1, we see that
ℏ(z)+zℏ′(z)1λℏ(z)≺1+(1−2μ)z1−z≺1+z1−z(λ=1+αkk(1−α);μ=k−αk−1k(1−α)). |
If we set λ1=1λ, γ2=0, C=1 and D=−1 in Lemma 4, then
ℏ(z)≺s(z)=1g(z)=z1λ(1−z)−2λ1/λ∫z0t(1/λ)−1(1−t)−2/λdt. |
By putting t=uz, and using Lemma 5, we obtain
ℏ(z)≺s(z)=1g(z)=11λ(1−z)2λ∫10u(1/λ)−1(1−uz)−2/λdu=[2G1(2λ,1,1λ+1;zz−1)]−1, |
which is the desired result of Theorem 4.
The present investigation was supported by the Key Project of Education Department of Hunan Province under Grant no. 19A097 of the P. R. China. The authors would like to thank the referees for their valuable comments and suggestions, which was essential to improve the quality of this paper.
The authors declare no conflict of interest.
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