On a Generalized Briot-Bouquet Differential Subordination
aa r X i v : . [ m a t h . C V ] J a n ON A GENERALIZED BRIOT-BOUQUET DIFFERENTIALSUBORDINATION
S. SIVAPRASAD KUMAR AND PRIYANKA GOEL ∗ Abstract.
We introduce and study the following special type of differential subordinationimplication: p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ h ( z ) ⇒ p ( z ) ≺ h ( z ) , (0.1)which involves generalization of the Briot-Bouquet differential subordination, where Q ( z )is analytic and 0 = β, α ∈ C . Further, some special cases of our result are also discussed.Finally, analogues of open door lemma and integral existence theorem with applications tounivalent functions are obtained. Introduction
Let H be the class of analytic functions defined on the open unit disc D := { z ∈ C : | z | < } . For any positive integer n and complex number a, H [ a, n ] denotes the subclass of H consisting of functions of the form f ( z ) = a + a n z n + a n +1 z n +1 + · · · . Let A n be theclass of functions of the form f ( z ) = z + a n +1 z n +1 + · · · and denote A := A . The subclassof A consisting of univalent functions is denoted by S . A function f ∈ A is said to betypically real in D if for every non-real z in D , we have sign (Im f ( z )) = sign (Im z ) . Theclass of all such functions is denoted by
T R . Let P be the class of analytic functions ofthe form p ( z ) = 1 + c z + c z + · · · such that Re p ( z ) > z in D , P is known as theCarath´eodory class. Given two analytic functions f and F, we say that f is subordinateto F, denoted by f ≺ F if there exists an analytic function ω ( z ) , with ω (0) = 0 and | ω ( z ) | < , such that f ( z ) = F ( ω ( z )) . In particular, if F is univalent then f ≺ F if andonly if f (0) = F (0) and f ( D ) ⊂ F ( D ) . Assume R to be the class of functions with boundedturning consisting of all the functions f ∈ A such that Re f ′ ( z ) > z ∈ D ) , clearly R ⊂ S . A function f in S is said to be starlike if and only if Re( zf ′ ( z ) /f ( z )) > D and the class of starlike functions is denoted by S ∗ . Similarly, the class of convex functions,denoted by K consists of all those functions f in S for which Re(1 + zf ′′ ( z ) /f ′ ( z )) > D . Let S ∗ ( α ) (0 ≤ α <
1) be the subclass of S ∗ consisting of the functions f satisfyingRe( zf ′ ( z ) /f ( z )) > α. We say that a function f is strongly starlike of order γ (0 < γ ≤ , whenever | arg ( zf ′ ( z ) /f ( z )) | < γπ/ SS ∗ ( γ ) , note that SS ∗ (1) = S ∗ (0) = S ∗ . Coman [6] defined that a function f ∈ A is said to be almost strongly starlike of order α, α ∈ (0 , , with respect to the function Mathematics Subject Classification.
Key words and phrases.
Starlike functions, Briot Bouquet, Subordination.*Corresponding authorThe work presented here was supported by the Council of Scientific and Industrial Research(CSIR).Ref.No.:08/133(0018)/2017-EMR-I. g ∈ S ∗ (1 − α ) if g ( z ) f ′ ( z ) g ′ ( z ) f ( z ) ≺ (cid:18) z − z (cid:19) α or equivalently, (cid:12)(cid:12)(cid:12)(cid:12) arg g ( z ) f ′ ( z ) g ′ ( z ) f ( z ) (cid:12)(cid:12)(cid:12)(cid:12) < α π F -starlike functions, denoted by F S ∗ as F S ∗ = { f ∈ A : Re (cid:18) F ( z ) f ′ ( z ) F ′ ( z ) f ( z ) (cid:19) > } , where F is fixed univalent function on the closed unit disk D , with at most a single pole on ∂ D and F (0) = 0. In the year 1992, Ma and Minda [13] came up with a generalized subclassof S ∗ as well as K , defined using subordination. They considered an analytic function ϕ, having positive real part, which is starlike with respect to ϕ (0) = 1, symmetric about the realaxis and satisfies ϕ ′ (0) > D . Various subclasses of S ∗ are evolved for different choicesof ϕ in due course of time ( see [5, 9, 14, 20, 24]). Note that ϕ is a typically real functionas ϕ ′ (0) > . Miller and Mocanu [16] introduced the theory of differential subordination asan analogue to differential inequalities, using the concept of subordination. The gist of thiswhole theory is the following implication: ψ ( p ( z ) , zp ′ ( z ) , z p ′′ ( z )) ≺ h ( z ) ⇒ p ( z ) ≺ q ( z ) , z ∈ D , (1.1)where ψ ( p ( z ) , zp ′ ( z ) , z p ′′ ( z )) is analytic in D . Given a complex function ψ : C × D → C anda function h , univalent in D , if p is analytic in D and satisfies the differential subordination ψ ( p ( z ) , zp ′ ( z ) , z p ′′ ( z )) ≺ h ( z ) , (1.2)then p is called a solution (1.2). The univalent function q is called a dominant of the solutionsof the differential subordination, if p ≺ q for all p satisfying (1.2). A dominant ˜ q that satisfies˜ q ≺ q for all dominants q is said to be best dominant of (1.2). The implication (1.1) givesrise to three types of differential subordination problems stated in [16, Chapter.2] and agood deal of literature associated with them is available (see [1, 4, 8, 9, 11, 12, 17, 19]). Usingthis theory, a special type of first order differential subordination, known as Briot-Bouquetdifferential subordination defined by p ( z ) + zp ′ ( z ) βp ( z ) + α ≺ h ( z ) , (1.3)was studied by Miller and Mocanu [16]. Many implication results were proved later associ-ating (1.3). Ruschewyh and Singh [22] considered Briot-Bouquet differential subordinationin a more particular form given by p ( z ) + zp ′ ( z ) βp ( z ) + α ≺ z − z with α ≥ β > . Later it was generalized to the form given by (1.3), in which h ( z ) is taken to be a univalent function and α, β = 0 are extended to complex numbers.This particular differential subordination has vast number of applications in the univalentfunction theory, see [2, 6, 7, 15, 18, 23] and the references therein. Motivated by the abovework, in the present investigation, we consider the following differential subordination p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ h ( z ) , N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 3 which is clearly a generalization of the Briot-Bouquet differential subordination and findconditions on α, β and Q ( z ) so that the implication (0.1) holds. Further, we establishcertain subordination results analogous to open door lemma and integral existence theorem.Apart from deriving other similar results, we also find sufficient conditions for starlikenessand univalence as applications of our results. For a better understanding of our main results,one may refer to [16, Ch.2] for the prerequisites.2. Generalized Briot-Bouquet Differential Subordination
We present here all implication results pertaining to the proposed generalized Briot-Bouquet differential subordination. We begin with the following result:
Theorem 2.1.
Let h be convex in D and α, β ∈ C with β = 0 . If Q ∈ H [1 , n ] be such thatthe following conditions hold: ( i ) Re (cid:18) βh ( z ) + α (cid:19) > z ∈ D ) . ( ii ) Re (cid:18) βh ( ζ ) + α + ( Q ( z ) − h ( ζ ) ζ h ′ ( ζ ) (cid:19) > z ∈ D , ζ ∈ h − ( p ( D ))) , where D = { z ∈ D : p ( z ) = h ( ζ ) for some ζ ∈ ∂ D } . If p is analytic in D with p (0) = h (0) and p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ h ( z ) , (2.1) then p ( z ) ≺ h ( z ) . Proof.
Let us suppose p is not subordinate to h . Then by [16, Lemma 2.2d, pp.24] thereexists z ∈ D , ζ ∈ ∂ D and m ≥ p ( z ) = h ( ζ ) and z p ′ ( z ) = mζ h ′ ( ζ ) andtherefore we have ψ := ψ ( p ( z ) , z p ′ ( z )) = ψ ( h ( ζ ) , mζ h ′ ( ζ )) = h ( ζ ) Q ( z ) + mζ h ′ ( ζ ) βh ( ζ ) + α , which yields Re ψ − h ( ζ ) ζ h ′ ( ζ ) = Re (cid:18) ( Q ( z ) − h ( ζ ) ζ h ′ ( ζ ) + mβh ( ζ ) + α (cid:19) . Using the fact that m ≥ ψ − h ( ζ ) ζ h ′ ( ζ ) ≥ Re (cid:18) ( Q ( z ) − h ( ζ ) ζ h ′ ( ζ ) + 1 βh ( ζ ) + α (cid:19) > , which implies (cid:12)(cid:12)(cid:12)(cid:12) arg ψ − h ( ζ ) ζ h ′ ( ζ ) (cid:12)(cid:12)(cid:12)(cid:12) < π . Since h ( D ) is convex, h ( ζ ) ∈ h ( ∂ D ) and ζ h ′ ( ζ ) is the outward normal to h ( ∂ D ) at h ( ζ ) , we conclude that ψ / ∈ h ( D ) , which contradicts (2.1) and hence p ( z ) ≺ h ( z ) . (cid:3) Remark . If we take Q ( z ) = 1 in Theorem 2.1, it reduces to [16, Theorem 3.2a]. S. SIVAPRASAD KUMAR AND P. GOEL
Corollary 2.3.
Let h be convex in D and α, β ∈ C with β = 0 . If Q ∈ H [1 , n ] and p isanalytic in D with p (0) = h (0) = ( k − / , where k ≥ , be such that Re (cid:18) βh ( ζ ) + α (cid:19) > k | Q ( z ) − |− Re( Q ( z ) −
1) ( z ∈ D , ζ ∈ ∂ D ) , (2.2) then p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ h ( z ) ⇒ p ( z ) ≺ h ( z ) . Proof.
Since k ≥ , from (2.2) it is clear that for ζ ∈ ∂ D , Re (cid:18) βh ( ζ ) + α (cid:19) > . (2.3)Since h is convex, the above inequality holds on D as well. Also, we can say that ˜ h ( z ) := h ( z ) − h (0) ∈ C . Using Marx Strohh˜acker theorem [16], we have Re( ζ ˜ h ′ ( ζ ) / ˜ h ( ζ )) > / , which is equivalent to (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ h ( ζ ) ζ ˜ h ′ ( ζ ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ , which implies (cid:12)(cid:12)(cid:12)(cid:12) h ( ζ ) ζ h ′ ( ζ ) − (cid:12)(cid:12)(cid:12)(cid:12) ≤ | h (0) || h ′ ( ζ ) | . (2.4)Since ˜ h ∈ C , we have | ˜ h ′ ( z ) |≥ / (1 + r ) on | z | = r [10, Theorem.9, pp 118]. We know that ζ ∈ ∂ D , so we have | h ′ ( ζ ) | = | ˜ h ′ ( ζ ) |≥ / . Thus (2.4) reduces to (cid:12)(cid:12)(cid:12)(cid:12) h ( ζ ) ζ h ′ ( ζ ) − (cid:12)(cid:12)(cid:12)(cid:12) ≤ k. Note that if
X, Y ∈ C and | X − |≤ K, thenRe( X.Y ) = Re Y + Re Y ( X − ≥ Re Y − | Y | K. Using this inequality, we can say thatRe (cid:18) ( Q ( z ) − h ( ζ ) ζ h ′ ( ζ ) + 1 βh ( ζ ) + α (cid:19) ≥ Re( Q ( z ) − − k | Q ( z ) − | + Re (cid:18) βh ( ζ ) + α (cid:19) , which by using (2.2) impliesRe (cid:18) ( Q ( z ) − h ( ζ ) ζ h ′ ( ζ ) + 1 βh ( ζ ) + α (cid:19) > . (2.5)From (2.3) and (2.5), we may conclude that the conditions (i) and (ii) of Theorem 2.1 aresatisfied and as its application, the result follows. (cid:3) Corollary 2.4.
Let Q ∈ H [1 , be a function such that | Q ( z ) |≤ M ( z ∈ D ) for some M > and α, β ∈ C with β = 0 . Suppose p is analytic and h is convex in D with p (0) = h (0) = 1 such that Re (cid:18) βh ( ζ ) + α (cid:19) > M + 1) ( ζ ∈ ∂ D ) , (2.6) N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 5 then p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ h ( z ) ⇒ p ( z ) ≺ h ( z ) . Proof.
We know that − Re( Q ( z ) − ≤ | Q ( z ) − | and since p (0) = 1 , in view of Corollary 2.3,we have k = 5 . Thus k | Q ( z ) − |− Re( Q ( z ) − ≤ | Q ( z ) − |≤ | Q ( z ) | +1) ≤ M + 1) . (2.7)Since (2.2) holds due to (2.6) and (2.7) and therefore the result follows from Corollary 2.3. (cid:3) Corollary 2.5.
Let h be convex in D with h (0) = 1 and α, β ∈ C with β = 0 . Let g ∈ A bedefined as g ( z ) = z exp Z z Q ( t ) − t dt, (2.8) such that h and Q satisfy the conditions ( i ) and ( ii ) of Theorem 2.1. If f ∈ A and F isgiven by F ( z ) = I [ f, g ] = (cid:18) α + βg α ( z ) Z z g ′ ( t ) g α − ( t ) f β ( t ) dt (cid:19) /β , (2.9) then zf ′ ( z ) f ( z ) ≺ h ( z ) ⇒ zF ′ ( z ) /F ( z ) zg ′ ( z ) /g ( z ) ≺ h ( z ) . Proof.
From (2.8), we have Q ( z ) = zg ′ ( z ) /g ( z ) and let us suppose p ( z ) = zF ′ ( z ) / ( Q ( z ) F ( z )) . Then by differentiating (2.9) and appropriately replacing the expressions, we have p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α = zf ′ ( z ) f ( z ) . Since zf ′ ( z ) /f ( z ) ≺ h ( z ) , the result now follows from Theorem 2.1. (cid:3) If we take h ( z ) = ((1+ z ) / (1 − z )) γ with γ ∈ (0 ,
1] in Corollary 2.5, we obtain the followingresult.
Corollary 2.6.
Let f ∈ A and g, Q and F are as defined in (2.8) and (2.9) respectivelysuch that Re( Q ( z ) − > − γ . Then f ∈ SS ∗ ( γ ) ⇒ F is almost strongly starlike of order γ w.r.t the function g. If we take h ( z ) = (1 + z ) / (1 − z ) in Corollary 2.5, we obtain the following result. Corollary 2.7.
Let f ∈ A and g, Q and F are as defined in (2.8) and (2.9) respectivelysuch that f ∈ S ∗ and g is univalent on D , then F is a g -starlike function. Now we list some of the special cases of Theorem 2.1 here below:
Corollary 2.8.
Let Q ∈ H [1 , and α, β be non-negative real numbers with β = 0 suchthat | Q ( z ) − | < / ( βe + α ) on D . Suppose p is analytic with p (0) = 1 , then p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ e z ⇒ p ( z ) ≺ e z . S. SIVAPRASAD KUMAR AND P. GOEL
Proof.
If we take h ( z ) = e z , then for α ≥ β > , we haveRe (cid:18) βe z + α (cid:19) ≥ βe + α > . (2.10)Further, we observe for z ∈ D and ζ ∈ ∂ D , Re (cid:18) Q ( z ) − ζ + 1 βe z + α (cid:19) ≥ −| Q ( z ) − | + 1 βe + α > . (2.11)From (2.10) and (2.11), we may conclude that both the conditions of Theorem 2.1 aresatisfied and thus the result now follows from Theorem 2.1. (cid:3) Corollary 2.9.
Let Q ∈ H [1 , and α ≥ , β > be such that | Q ( z ) − | < Re( Q ( z ) −
1) + 12( √ β + α ) . (2.12) Suppose p is analytic with p (0) = 1 , then p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ √ z ⇒ p ( z ) ≺ √ z. Proof.
Let h ( z ) = √ z, then condition (i) of Theorem 2.1 is satisfied clearly as for α ≥ β > , Re (cid:18) β √ z + α (cid:19) ≥ β √ α > , z ∈ D . Now for z ∈ D and ζ ∈ ∂ D , we have from (2.12)Re (cid:18) Q ( z ) − (cid:18) ζ (cid:19) + 1 β √ z + α (cid:19) ≥ Q ( z ) − − | Q ( z ) − | + 1 √ β + α> , which implies that condition (ii) of Theorem 2.1 is satisfied. Thus the result follows at oncefrom Theorem 2.1. (cid:3) Corollary 2.10.
Let Q ∈ H [1 , be such that Q ′ (0) > and α, β be non-negative realnumbers with β = 0 . Suppose p is analytic with p (0) = 1 and p ′ (0) > , then for < γ ≤ ,p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ (cid:18) z − z (cid:19) γ ⇒ p ( z ) ≺ (cid:18) z − z (cid:19) γ . Proof.
Let h ( z ) = ((1 + z ) / (1 − z )) γ . Then condition (i) of Theorem 2.1 clearly holds. Forcondition (ii) to hold, we need to show thatRe ( Q ( z ) −
1) 1 − ζ γζ + 1 β (cid:18) ζ − ζ (cid:19) γ + α > z ∈ D , ζ ∈ ∂ D ) . Let Q ( z ) = 1 + a z + a z + · · · and define R ( z ) = Q ( z ) − , then R (0) = 0 . Since Q ′ (0) > , it is typically real and it is easy to conclude that R ( z ) is typically real. Weknow that ζ ∈ h − ( p ( D )) , where D := { z ∈ D : p ( z ) = h ( ζ ) for some ζ ∈ D } . Since
N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 7 h ( z ) = ((1 + z ) / (1 − z )) γ is typically real and p ′ (0) > , we have sign (Im z ) = sign (Im ζ ) . Now we consider Re (cid:18) ( Q ( z ) −
1) 1 − ζ γζ (cid:19) = 12 γ (cid:18) Re( Q ( z ) −
1) Re (cid:18) − ζ ζ (cid:19) − Im( Q ( z ) −
1) Im (cid:18) − ζ ζ (cid:19)(cid:19) = − γ Im( Q ( z ) −
1) Im (cid:18) − ζ ζ (cid:19) . Taking ζ = e iθ (0 ≤ θ ≤ π ) , we have − Im (cid:18) − ζ ζ (cid:19) = 2 sin θ ( > , θ ∈ (0 , π ) ,< , θ ∈ ( π, π ) . Since Q ( z ) − sign (Im( Q ( z ) − sign (Im z ) . Thus we haveRe (cid:18) ( Q ( z ) −
1) 1 − ζ γζ (cid:19) = 2 sin θ Im( Q ( z ) − ≥ . Also for α ≥ β > , we have Re (cid:18) β (cid:18) z − z (cid:19) γ + α (cid:19) > ( Q ( z ) −
1) 1 − ζ γζ + 1 β (cid:18) ζ − ζ (cid:19) γ + α > . Therefore the result follows at once from Theorem 2.1. (cid:3)
Now we use a different technique to prove the next two results, which demonstrates thesimilar implication for Janowski functions.
Theorem 2.11.
Let p ( z ) be analytic in D with p (0) = 1 and Q ∈ H [1 , n ] be a function suchthat | Q ( z ) | < M for some M > . Let − ≤ B < A < and − < E < D ≤ satisfy ( A − B )(1 − A )(1 + E ) > (1 + | A | )( β + α + | βA + αB | )((1 + D )(1 − B )+ M (1 + E )(1 − A )) (2.13) for some α and β, where α + β > . If p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ Dz Ez , then p ( z ) ≺ (1 + Az ) / (1 + Bz ) . Proof.
Let us define P ( z ) and ω ( z ) as P ( z ) := p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α , ω ( z ) := p ( z ) − A − Bp ( z ) . Then ω ( z ) is meromorphic in D and ω (0) = 0 . By the definition of P ( z ) and ω ( z ) , we have P ( z ) = 1 + Aω ( z )1 + Bω ( z ) Q ( z ) + ( A − B ) zω ′ ( z )(1 + Bω ( z ))[ β (1 + Aω ( z )) + α (1 + Bω ( z ))] . S. SIVAPRASAD KUMAR AND P. GOEL
Now we need to show that | ω ( z ) | < D . On the contrary, let us assume that there existsa point z ∈ D such that max | z |≤| z | | ω ( z ) | = | ω ( z ) | = 1 . Then by [21, Lemma 1.3, pp.28], there exists k ≥ z ω ′ ( z ) = kω ( z ) . Now bytaking ω ( z ) = e iθ (0 ≤ θ ≤ π ) , we have | P ( z ) | = (cid:12)(cid:12)(cid:12)(cid:12) Aω ( z )1 + Bω ( z ) Q ( z ) + ( A − B ) kω ( z )(1 + Bω ( z ))( β (1 + Aω ( z )) + α (1 + Bω ( z ))) (cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12) Ae iθ Be iθ (cid:12)(cid:12)(cid:12)(cid:12) (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ( A − B ) k (1 + Ae iθ )( β (1 + Ae iθ ) + α (1 + Be iθ )) (cid:12)(cid:12)(cid:12)(cid:12) − | Q ( z ) | (cid:19) . We know that for p > , | p + qe iθ | = p + q + 2 pq cos θ, attains its maximum at θ = 0 if q > θ = π if q ≤
0. So, max ≤ θ ≤ π | p + qe iθ | = ( p + | q | ) . Thus | P ( z ) |≥ − A − B (cid:18) ( A − B ) k (1 + | A | )( β + α + | βA + αB | ) − M (cid:19) . Clearly the expression on the right hand side is an increasing function of k and attains itsminimum at k = 1 . So | P ( z ) |≥ − A − B (cid:18) ( A − B )(1 + | A | )( β + α + | βA + αB | ) − M (cid:19) . Now from (2.13), we have | P ( z ) | > D E , which contradicts the fact that P ( z ) ≺ (1+ Dz ) / (1+ Ez ) and hence completes the proof. (cid:3) Using Theorem 2.11, we obtain the following corollary.
Corollary 2.12.
Let p ( z ) be analytic in D with p (0) = 1 and Q be a function such that z exp Z z Q ( t ) − t dt ∈ S ∗ ( φ ) . Let − ≤ B < A < , − < E < D ≤ and α, β be such that α + β > . Then p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ Dz Ez ⇒ p ( z ) ≺ Az Bz , whenever any of the following cases hold: ( i ) φ ( z ) = e z and ( A − B )(1 − A )(1 + E ) > (1 + | A | )( β + α + | βA + αB | )((1 + D )(1 − B ) + e (1 + E )(1 − A ))( ii ) φ ( z ) = √ z and ( A − B )(1 − A )(1 + E ) > (1 + | A | )( β + α + | βA + αB | )((1 + D )(1 − B ) + √ E )(1 − A ))( iii ) φ ( z ) = 2 / (1 + e − z ) and ( A − B )(1 − A )(1 + E )(1 + e ) > (1 + | A | )( β + α + | βA + αB | )((1 + D )(1 − B )(1 + e ) + 2 e (1 + E )(1 − A ))( iv ) φ ( z ) = 1 + ze z and ( A − B )(1 − A )(1 + E ) > (1 + | A | )( β + α + | βA + αB | )((1 + D )(1 − B ) + (1 + e )(1 + E )(1 − A ))( v ) φ ( z ) = z + √ z and ( A − B )(1 − A )(1 + E ) > (1 + | A | )( β + α + | βA + αB | )((1 + D )(1 − B ) + (1 + √ E )(1 − A )) . N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 9
Since the proof of Theorem 2.11 loses its validity when A = 1 or E = − , the theoremdoes not reduce to the case when P ( z ) and p ( z ) , both are subordinate to (1 + az ) / (1 − z )for 0 ≤ a ≤ . This case, we handle in the following result with a different approach.
Theorem 2.13.
Let α ≥ , β > , ≤ a ≤ . Assume p ( z ) to be analytic in D with p (0) = 1 and p ′ (0) > . Also, let Q ( z ) ∈ H [1 , be such that Q ′ (0) > and Re Q ( z ) < . If p satisfies p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α ≺ az − z , (2.14) then p ( z ) ≺ (1 + az ) / (1 − z ) . Proof.
We need to show that p ( z ) ≺ q ( z ) = (1 + az ) / (1 − z ) . For if p ⊀ q , then using [16,Lemma 2.2d, pp.24], there exists a point z = r e iθ ∈ D , ζ ∈ ∂ D \{ } and m ≥ p ( z ) = q ( ζ ) , z p ′ ( z ) = mζ q ′ ( ζ ) and p ( D r ) ⊂ q ( D ) . Since ζ is a boundary point,we may assume that ζ = e iθ for θ ∈ [0 , π ] . Then p ( z ) Q ( z ) + z p ′ ( z ) βp ( z ) + α = q ( ζ ) Q ( z ) + mζ q ′ ( ζ ) βq ( ζ ) + α = (cid:18) − a i a θ/ (cid:19) Q ( z ) − m ( a + 1) e iθ (1 − e iθ )(( β + α ) + ( βa − α ) e iθ ) . Taking Q ( z ) = u + iv, we have Re (cid:18) p ( z ) Q ( z ) + z p ′ ( z ) βp ( z ) + α (cid:19) = (1 − a ) u − (1 + a ) cot ( θ/ v − m (1 + a )( β (1 − a ) + 2 α )(1 − cos θ )( β (1 − a ) + 2 α ) (1 − cos θ ) + ( β ( a + 1) sin θ ) . Since Q ′ (0) >
0, it is clear that Q ( z ) is typically real and thus sign ( v ) = sign (Im( Q ( z ))) = sign (Im( z )) . (2.15)We also have p ′ (0) > q ( z ) = (1 + az ) / (1 − z ) is typically real, which implies sign (Im( z )) = sign ((Im( p ( z )))) = sign ((Im( q ( ζ )))) = sign (Im( ζ )) = sign (sin θ ) . (2.16) From (2.15) and (2.16), we obtain sign ( v ) = sign (sin θ ) , which is sufficient to concludethat v cot θ/ > θ ∈ [0 , π ] . Also we know that α ≥ , β > , m ≥ ≤ a ≤ . Therefore Re (cid:18) p ( z ) Q ( z ) + z p ′ ( z ) βp ( z ) + α (cid:19) < (1 − a ) u < − a , which contradicts (2.14) and hence the result. (cid:3) Other Subordination Results and Applications
We begin with the following analogue of open door lemma:
Theorem 3.1.
Let p ( z ) be analytic in D with p (0) = 1 and Q ( z ) ∈ P . Suppose that α ≥ , β > and p satisfies p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α = 1 , (3.1) then Re p ( z ) > . Proof.
Define the analytic function g by g ( z ) = z exp Z z Q ( t ) − t dt. Then one can verify that the function p ( z ) , given by p ( z ) = g α ( z ) z β β (cid:18)Z z g α − ( t ) g ′ ( t ) t β (cid:19) − − αβ (3.2)is analytic in D and p ∈ H [1 , n ] . The logarithmic differentiation of (3.2) reveals that p is asolution of the differential equation (3.1). We now use [16, Theorem 2.3i, pp.35] to provethat Re p ( z ) >
0. Let Ω = { } and ψ ( r, s ; z ) = rQ ( z ) + s/ ( βr + α ) , then (3.1) can bewritten as { ψ ( p ( z ) , zp ′ ( z ); z ) | z ∈ D } ⊂ Ω . In view of [16, Theorem 2.3i, pp.35], it is sufficient to show that ψ ∈ Ψ n [Ω , , which meansadmissibility conditions defined for the class Ψ n [Ω ,
1] are satisfied by ψ or equivalently: ψ ( ρi, σ ; z ) = σβρi + α + Q ( z ) ρi = 1 , (3.3)where ρ ∈ R , σ ≤ − n ρ ) , z ∈ D and n ≥ . Suppose on the contrary if we assume (3.3)is false, then there exists some values ρ , σ and z such that σ βρ i + α + iρ Q ( z ) = 1 . (3.4)If we let Q ( z ) = S ( z ) + iT ( z ) , then (3.4) yields σ αα + β ρ − ρ T ( z ) = 1 and − βσ ρ α + β ρ + ρ S ( z ) = 0 . (3.5)Since σ < α ≥ , we have ρ = 0 . Therefore from (3.5), we deduce thatRe Q ( z ) = S ( z ) = βσ α + β ρ ≤ − βn (1 + ρ )2( α + β ρ ) < , which contradicts the hypothesis and hence it follows that Re p ( z ) > . (cid:3) Corollary 3.2.
Let f ∈ A be such that f ( z ) /z = 0 in D and f ∈ R . Then Re f ( z ) /z > . Proof.
Let Q ( z ) = f ′ ( z ) then Q ∈ P as f ∈ R . Choose p ( z ) = z/f ( z ) , β = 1 and α = 0,then clearly p (0) = 1 and p ( z ) satisfies (3.1). Now by an application of Theorem 3.1, itfollows that Re p ( z ) > (cid:3) Here below we consider an analogue of integral existence theorem:
N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 11
Theorem 3.3.
Let ϕ, φ ∈ H [1 , n ] , with ϕ ( z ) φ ( z ) = 0 in D . Let λ, η, γ and δ be complexnumbers with η = 0 , λ + δ = η + γ = 1 and α, β be non negative real numbers with β = 0 .Let g ∈ A n and suppose that Q ( z ) = λ zg ′ ( z ) g ( z ) + zϕ ′ ( z ) ϕ ( z ) + δ ≺ z − z . (3.6) If F is defined by F ( z ) = (cid:18) α + βφ β ( z ) g λα ( z ) ϕ α ( z ) z δα + βγ Z z g λα ( t ) ϕ α ( t ) t β + δα − Q ( t ) dt (cid:19) ηβ , (3.7) then F ∈ A n , F ( z ) /z = 0 and for z ∈ D , Re η zF ′ ( z ) F ( z ) + zφ ′ ( z ) φ ( z ) + γλ zg ′ ( z ) g ( z ) + zϕ ′ ( z ) ϕ ( z ) + δ > . (3.8) Proof.
Let p ( z ) be given by p ( z ) = 1 β g λα ( z ) ϕ α ( z ) z β + δα (cid:18)Z z g λα ( t ) ϕ α ( t ) t β + δα − Q ( t ) dt (cid:19) − − αβ . (3.9)Then p ( z ) = 1 + p n z n + p n +1 z n +1 + · · · is analytic in D and p ∈ H [1 , n ] . The logarithmic differentiation of (3.9) shows that p ( z )satisfies (3.1) with Q ( z ) as given in (3.6). From hypothesis, we have Q ( z ) ≺ (1 + z ) / (1 − z ) . Thus the hypothesis of the Theorem. 3.1 is fulfilled by p as well as Q and hence it followsthat Re p ( z ) > . Using (3.7) and (3.9), we get F ( z ) = (cid:18) ( α + β ) z β (1 − γ ) φ β ( z )( βp ( z ) + α ) (cid:19) ηβ = z (cid:18) α + βφ β ( z )( βp ( z ) + α ) (cid:19) ηβ . (3.10)Clearly the expression in the bracket is analytic and non-zero, so we deduce that F ∈ A n and F ( z ) /z = 0 . By differentiating (3.10) logarithmically, we obtain η zF ′ ( z ) F ( z ) = 1 − γ − zφ ′ ( z ) φ ( z ) − zp ′ ( z ) βp ( z ) + γ , which by using (3.1) simplifies to η zF ′ ( z ) F ( z ) + zφ ′ ( z ) φ ( z ) + γ = p ( z ) Q ( z ) . Therefore Re 1 Q ( z ) (cid:18) η zF ′ ( z ) F ( z ) + zφ ′ ( z ) φ ( z ) + γ (cid:19) > , which is equivalent to (3.8) and hence completes the proof. (cid:3) We now obtain the following special case of the above theorem when δ = γ = 0 , λ = η = 1and ϕ ( z ) = 1 = φ ( z ) . Corollary 3.4.
Let g ∈ S ∗ , α ≥ , β > and F is defined as F ( z ) = α + βg α ( z ) Z z g ′ ( t ) g α − ( t ) t β dt, then F is almost strongly starlike with respect to g and hence univalent. Theorem 3.5.
Let f, g ∈ A n and φ ∈ H [1 , n ] with φ (0) = 1 and φ ( z ) = 0 in D . Let β, α and σ be complex numbers such that β = 0 and Re( βh ( z ) + α ) > , where h is a convexfunction with h (0) = 1 in D . Suppose β zf ′ ( z ) f ( z ) + σ zg ′ ( z ) g ( z ) ≺ βh ( z ) + σ. (3.11) If F is defined by F ( z ) = (cid:18) β + αz α φ ( z ) Z z f β ( t ) g σ ( t ) t α − σ − dt (cid:19) β , (3.12) then F ( z ) ∈ A n and zF ′ ( z ) F ( z ) + 1 β zφ ′ ( z ) φ ( z ) ≺ h ( z ) . (3.13) Proof.
Let p ( z ) be given by p ( z ) = 1 β f β ( z ) g σ ( z ) z α − σ (cid:18)Z z f β ( t ) g σ ( t ) t α − σ − dt (cid:19) − − αβ . (3.14)Then p ( z ) is analytic in D and p (0) = 1 . Upon logarithmic differentiation of (3.14), wededuce that p ( z ) satisfies p ( z ) + zp ′ ( z ) βp ( z ) + α = zf ′ ( z ) f ( z ) + σβ (cid:18) zg ′ ( z ) g ( z ) − (cid:19) . (3.15)From (3.11) and (3.15), it can be concluded that p ( z ) + zp ′ ( z ) βp ( z ) + α ≺ h ( z ) . By applying [16, Theorem 3.2a, pp.81], we obtain p ( z ) ≺ h ( z ) . Substituting (3.12) in (3.14),we obtain p ( z ) = 1 β (cid:18) ( α + β ) f β ( z ) g σ ( z ) F β ( z ) φ ( z ) z σ − α (cid:19) . Differentiating logarithmically the following equation βp ( z ) + α = ( α + β ) f β ( z ) g σ ( z ) F β ( z ) φ ( z ) z σ and using (3.15), we obtain p ( z ) = zF ′ ( z ) F ( z ) + 1 β zφ ′ ( z ) φ ( z )and thus (3.13) follows. (cid:3) We now derive another result analogous to open door lemma as follows:
N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 13
Lemma 3.6.
Let n be a positive integer and α, β be non-negative real numbers with β = 0 .Let Q ∈ H [1 , n ] satisfy Q ( z ) ≺ z + nzβ + α (1 + z ) ≡ h ( z ) . (3.16) If p ∈ H [1 , n ] satisfies the differential equation p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α = 1 , (3.17) then p ( z ) ≺ / (1 + z ) . Proof.
Let us set q ( z ) = 1 / (1 + z ) , then h ( z ) = 1 q ( z ) − nzq ′ ( z ) q ( z )( βq ( z ) + α ) . If g ( z ) = z/ ( β + α (1 + z )) , then we haveRe zg ′ ( z ) g ( z ) = 1 − Re αzβ + α (1 + z ) > − αβ + 2 α = β + αβ + 2 α > . Since g is starlike andRe zh ′ ( z ) g ( z ) = Re (cid:18) ( β + α (1 + z )) (cid:18) n ( β + α )( β + α (1 + z )) (cid:19)(cid:19) = β + α (1 + Re z ) + n ( β + α ) Re (cid:18) β + α (1 + z ) (cid:19) ≥ β + n ( β + α ) β + 2 α > , we deduce that h is close to convex and hence univalent in D . Now we consider the boundarycurve of h defined as w = h ( e iθ ) = u ( θ ) + iv ( θ ) , θ ∈ ( − π, π ) . Suppose r ( θ ) = | h ( e iθ ) | = | e − iθ h ( e iθ ) | = (cid:12)(cid:12)(cid:12)(cid:12) e − iθ + 1 + nβ + α (1 + e iθ ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) θ + n ( β + α (1 + cos θ )) d ( θ ) − i (cid:18) sin θ + nα sin θd ( θ ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) , (3.18)where d ( θ ) = ( β + α (1 + cos θ )) + ( α sin θ ) . After simplifying (3.18), we have r ( θ ) = s θ ) + n + 2 n (2 α + β )(1 + cos θ ) d ( θ ) . By using (3.16) and (3.17), we deduce that Q ( z ) = 1 p ( z ) − zp ′ ( z ) p ( z )( βp ( z ) + α ) ≺ h ( z ) . (3.19) On the contrary, if p ⊀ q then by [16, Lemma 2.2d], there exist points z ∈ D and ζ ∈ ∂ D and m ≥ n, such that p ( z ) = q ( ζ ) and z p ′ ( z ) = mζ q ′ ( ζ ) . From (3.19), we have Q ( z ) = 1 q ( ζ ) − mζ q ′ ( ζ ) q ( ζ )( βq ( ζ ) + α ) = 1 + ζ + mζ β + α (1 + ζ ) . For ζ = e iθ , we have | Q ( z ) | = s θ ) + m + 2 m (2 α + β )(1 + cos θ ) d ( θ ) ≥ r ( θ ) θ ∈ ( − π, π ) , where r ( θ ) is given in (3.18). This implies that Q ( z ) / ∈ h ( D ) , which is a contradiction andthus p ( z ) ≺ / (1 + z ) . (cid:3) Theorem 3.7.
Let n be a positive integer and α, β be non negative real numbers with β = 0 .Let f ∈ A n and F = I α,β [ f ] be defined as I α,β [ f ] = (cid:18) α + βf α ( z ) Z z f α − ( t ) f ′ ( t ) t β dt (cid:19) /β . (3.20) If zf ′ ( z ) f ( z ) ≺ z + nzβ + α (1 + z ) , then | zf ′ ( z ) /f ( z ) | < | zF ′ ( z ) /F ( z ) | . Proof.
Let f ∈ A satisfy (3.20) and define p ( z ) = z β f α ( z ) (cid:18) β Z z f α − ( t ) f ′ ( t ) t β dt (cid:19) − − αβ . By the series expansion, it is easy to verify that p is well defined and p ∈ H [1 , n ] . If we let Q ( z ) = zf ′ ( z ) /f ( z ) , then it is easy to show that p satisfies (3.17). Hence by Lemma 3.6 wededuce that p ( z ) ≺ / (1 + z ) . Since p ( z ) = 0 , we can define the analytic function F ∈ A n by F ( z ) = z (cid:18) α + ββp ( z ) + α (cid:19) /β . A simple calculation shows that this function coincides with the function given in (3.20).So we obtain F ( z ) Q ( z ) zF ′ ( z ) = 1 p ( z ) ≺ z, which further implies (cid:12)(cid:12)(cid:12)(cid:12) zf ′ ( z ) /f ( z ) zF ′ ( z ) /F ( z ) − (cid:12)(cid:12)(cid:12)(cid:12) < (cid:3) Theorem 3.8.
Let n be a positive integer and α, β be non-negative real numbers with β = 0 .Let f ∈ A satisfy f ( z ) zf ′ ( z ) − (cid:18) zf ′′ ( z ) f ′ ( z ) + 1 − zf ′ ( z ) f ( z ) (cid:19) (cid:18) β zf ′ ( z ) f ( z ) + α (cid:19) − ≺ z + nzβ + α (1 + z ) . then zf ′ ( z ) /f ( z ) ≺ / (1 + z ) . N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 15
Proof.
Let p ( z ) = zf ′ ( z ) /f ( z ) . Then Q ( z ) = f ( z ) zf ′ ( z ) − (cid:18) zf ′′ ( z ) f ′ ( z ) + 1 − zf ′ ( z ) f ( z ) (cid:19) (cid:18) β zf ′ ( z ) f ( z ) + α (cid:19) − satisfies the following differential equation p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α = 1 . Now applying Lemma 3.6, we obtain p ( z ) ≺ / (1 + z ) and that completes the proof. (cid:3) If we take β = 1 , α = 0 and n = 1 in the above theorem, we obtain the following corollary,which is a particular case of a result of Tuneski [25, Theorem 2.5]. Corollary 3.9.
Let f ∈ A satisfies (cid:12)(cid:12)(cid:12)(cid:12) f ( z ) f ′′ ( z )( f ′ ( z )) (cid:12)(cid:12)(cid:12)(cid:12) < , then zf ′ ( z ) /f ( z ) ≺ / (1 + z ) . Theorem 3.10.
Let n be a positive integer and α, β be non-negative real numbers with β = 0 . Let f ∈ A satisfy f ( z ) z + (cid:18) zf ′ ( z ) f ( z ) − (cid:19) (cid:18) βzf ( z ) + α (cid:19) − ≺ z + nzβ + α (1 + z ) , then f ( z ) /z ≺ z. Proof.
Taking p ( z ) = z/f ( z ) and Q ( z ) = f ( z ) z + (cid:18) zf ′ ( z ) f ( z ) − (cid:19) (cid:18) βzf ( z ) + α (cid:19) − , the proof goes similarly as that of Theorem 3.8 and the result follows at once. (cid:3) By taking β = 1 , α = 0 and n = 1 in Theorem 3.10, we obtain the following result. Corollary 3.11.
Let f ∈ A be such that f ′ ( z ) ≺ z, then f ( z ) /z ≺ z, or equivalently | f ′ ( z ) − | < ⇒ | f ( z ) /z − | < . We now prove the following lemma in order to derive some sufficient conditions for star-likeness:
Lemma 3.12.
Let n be a positive integer and α, β be non negative integers with β = 0 . Suppose that either α < β < α or β < α < β and Q ∈ H [1 , n ] satisfy Q ( z ) ≺ z − z + 2 nz (1 − z )(( α + β ) + ( α − β ) z ) = h ( z ) . (3.21) If p ∈ H [1 , n ] satisfies the differential equation p ( z ) Q ( z ) + zp ′ ( z ) βp ( z ) + α = 1 , (3.22)then p ( z ) ≺ (1 − z ) / (1 + z ) . Proof.
Let us set q ( z ) = (1 − z ) / (1 + z ) , then h ( z ) = 1 q ( z ) − nzq ′ ( z ) q ( z )( βq ( z ) + α ) . We know that the Koebe function k ( z ) = z/ (1 − z ) is starlike andRe zh ′ ( z ) k ( z ) = Re (cid:18) (1 − z ) (cid:18) − z ) + 2 n (( α + β ) + ( α − β ) z )(1 − z ) (( α + β ) + ( α − β ) z ) (cid:19)(cid:19) = Re (cid:18) n (( α + β ) + ( α − β ) z )(( α + β ) + ( α − β ) z ) (cid:19) = 2 + 2 n Re (cid:18) ( α + β ) + ( α − β ) z (( α + β ) + ( α − β ) z ) (cid:19) . Now taking α + β = a, α − β = b and z = e iθ for θ ∈ ( − π, π ) , we have (cid:12)(cid:12)(cid:12)(cid:12) arg (cid:18) a + be iθ ( a + be iθ ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = | arg ( a + be iθ ) − a + be iθ ) | = (cid:12)(cid:12)(cid:12)(cid:12) arctan (cid:18) b sin 2 θa + b cos 2 θ (cid:19) − (cid:18) b sin θa + b cos θ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) arctan (cid:18) b sin 2 θa + b cos 2 θ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) + 2 (cid:12)(cid:12)(cid:12)(cid:12) arctan (cid:18) b sin θa + b cos θ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) . Since arctan x is an increasing function in ( − π, π ), we now find the maximum of g ( x ) := b sin x/ ( a + b cos x ) for x ∈ ( − π, π ) . Clearly, a > g ( x ) attains its maximum at x = π − arccos ( b/a ) when b > x = − arccos ( − b/a ) when b < , b/ √ a − b and − b/ √ a − b are the corresponding maximumvalues. Thus we have (cid:12)(cid:12)(cid:12)(cid:12) arg (cid:18) a + be iθ ( a + be iθ ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) | b |√ a − b (cid:19) . For the case when β < α < β, we observe that3 arctan (cid:18) | b |√ a − b (cid:19) = 3 arctan (cid:18) α − β √ αβ (cid:19) < (cid:18) r αβ (cid:19) < (cid:18) √ (cid:19) = π . The other case can also be verified in the similar way and it can be concluded that (cid:12)(cid:12)(cid:12)(cid:12) arg (cid:18) ( α + β ) + ( α − β ) z (( α + β ) + ( α − β ) z ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) < π , which further implies that Re( zh ′ ( z ) /k ( z )) > . So h is close to convex and hence univalentin D . Now we consider the boundary curve of h defined as h ( e iθ ) = u ( θ ) + iv ( θ ) , θ ∈ ( − π, π ) . Since e iθ is a boundary point, without loss of generality we may assume that (1 + e iθ ) / (1 − e iθ ) = iγ and thus r ( θ ) = | h ( e iθ ) | = (cid:12)(cid:12)(cid:12)(cid:12) iγ + 2 ne iθ (1 − e iθ )( β/iγ + α ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) iγ − n sin θ ( β/γ + iα ) (cid:12)(cid:12)(cid:12)(cid:12) . N A GENERALIZED BRIOT-BOUQUET DIFFERENTIAL SUBORDINATION 17 So r ( θ ) = s γ + n + 2 nαγ sin θ sin θ (( β/γ ) + α ) = s γ + n + 4 nαγ / (1 + γ )sin θ (( β/γ ) + α ) . (3.23)From (3.21) and (3.22), we deduce that Q ( z ) = 1 p ( z ) − zp ′ ( z ) p ( z )( βp ( z ) + α ) ≺ h ( z ) . (3.24)On the contrary if p is not subordinate to q , then by [16, Lemma.2.2d], there exist points z ∈ D and ζ ∈ ∂ D and m ≥ n, such that p ( z ) = q ( ζ ) and z p ′ ( z ) = mζ q ′ ( ζ ) . From (3.24), we have Q ( z ) = 1 q ( ζ ) − mζ q ′ ( ζ ) q ( ζ )( βq ( ζ ) + α ) = 1 + ζ − ζ + 2 mζ (1 − ζ )(( β + α ) + ( α − β ) ζ ) . For ζ = e iθ , we have | Q ( z ) | = s γ + m + 4 mαγ / (1 + γ )sin θ (( β/γ ) + α ) ≥ r ( θ ) θ ∈ ( − π, π ) , where r ( θ ) is given in (3.23). This implies that Q ( z ) / ∈ h ( D ) , which is a contradiction andthus we have p ( z ) ≺ (1 − z ) / (1 + z ) . (cid:3) Theorem 3.13.
Let n be a positive integer and α, β be non negative real numbers with β = 0 , either α < β < α or β < α < β. Let f ∈ A n and F = A α,β [ f ] is as definedin (3.20) . If zf ′ ( z ) f ( z ) ≺ z − z + 2 nz (1 − z )(( α + β ) + ( α − β ) z ) , then Re (cid:18) zF ′ ( z ) /F ( z ) zf ′ ( z ) /f ( z ) (cid:19) > . The proof of Theorem 3.13 follows by an application of Lemma 3.12, similar to that ofTheorem 3.7 and therefore it is omitted here.
Theorem 3.14.
Let n be a positive integer and α, β be non-negative real numbers with β = 0 . Suppose that either α < β < α or β < α < β and f ∈ A satisfies Θ( f ) ≺ z − z + 2 nz (1 − z )(( α + β ) + ( α − β ) z ) , where Θ( f ) = f ( z ) zf ′ ( z ) − (cid:18) zf ′′ ( z ) f ′ ( z ) + 1 − zf ′ ( z ) f ( z ) (cid:19) (cid:18) β zf ′ ( z ) f ( z ) + α (cid:19) − , then zf ′ ( z ) /f ( z ) ≺ (1 − z ) / (1 + z ) . The proof is omitted here as it is much akin to Theorem 3.8 and can be easily done byusing Lemma 3.12.
Remark . If we take n = 1 , β = 1 and α = 0 in Theorem 3.14, it reduces to a result ofTuneski [25, Theorem 2.1]. Theorem 3.16.
Let n be a positive integer and α, β be non-negative real numbers with β = 0 . Suppose that either α < β < α or β < α < β and f ∈ A satisfies f ( z ) z + (cid:18) zf ′ ( z ) f ( z ) − (cid:19) (cid:18) βzf ( z ) + α (cid:19) − ≺ z − z + 2 nz (1 − z )(( α + β ) + ( α − β ) z ) , then f ( z ) /z ≺ (1 + z ) / (1 − z ) . Proof.
Taking p ( z ) = z/f ( z ) and Q ( z ) = f ( z ) z + (cid:18) zf ′ ( z ) f ( z ) − (cid:19) (cid:18) βzf ( z ) + α (cid:19) − , the result follows by an application of Lemma 3.12. (cid:3) By taking β = 1 , α = 0 and n = 1 , we obtain the following result Corollary 3.17.
Let f ∈ A be such that f ′ ( z ) ≺ z − z + 2 z (1 − z ) , then f ( z ) z ≺ z − z . Concluding Remarks
The work carried out in this article is not an extension of any earlier work but deals withnew ideas and therefore there will be no parallel comparisons. As a matter of future scope,these ideas of transforming differential subordinations into integral operators and vice versacan be extended for other similar subordination cases too. Note that the lemmas obtainedin this paper are not confined to finding some starlikeness or univalence criteria alone, butalso paves way in establishing some exceptional implication results.
Funding
The present work is supported by the Council of Scientific and Industrial Research(CSIR).Ref.No.:08/133(0018)/2017-EMR-I
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