Briot-Bouquet differential subordination and Bernardi's integral operator
aa r X i v : . [ m a t h . C V ] J a n BRIOT–BOUQUET DIFFERENTIAL SUBORDINATION ANDBERNARDI’S INTEGRAL OPERATOR
KANIKA SHARMA, RASOUL AGHALARY, AND V. RAVICHANDRAN
Dedicated to Prof. Dato’ Indera Rosihan M. Ali
Abstract.
The conditions on A , B , β and γ are obtained for an analytic function p defined on the open unit disc D and normalized by p (0) = 1 to be subordinate to(1 + Az ) / (1 + Bz ), − ≤ B < A ≤ p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) is subordinate to e z . The conditions on these parameters are derived for the function p to be subordinateto √ z or e z when p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) is subordinate to (1 + Az ) / (1 + Bz ).The conditions on β and γ are determined for the function p to be subordinate to e z when p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) is subordinate to √ z . Related result for the function p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) to be in the parabolic region bounded by the Re w = | w − | isinvestigated. Sufficient conditions for the Bernardi’s integral operator to belong to thevarious subclasses of starlike functions are obtained as applications. introduction Let H denote the class of analytic functions in the unit disc D . For a natural number n , let H [ a, n ] be the subset of H consisting of functions p of the form p ( z ) = a + p n z n + p n +1 z n +1 + · · · . Suppose that h is a univalent function defined on D with h (0) = a andthe function p ∈ H [ a, n ]. The Briot–Bouquet differential subordination is the first orderdifferential subordination of the form(1.1) p ( z ) + zp ′ ( z ) βp ( z ) + γ ≺ h ( z ) , where β = 0 , γ ∈ C . This particular differential subordination has many interestingapplications in the theory of univalent functions. Ruschewyh and Singh [24] proved thatif the function p ∈ H [1 , β > , Re γ ≥ h ( z ) = (1 + z ) / (1 − z ) in (1.1) and thefunction q ∈ H satisfy the differential equation q ( z ) + zp ′ ( z ) βp ( z ) + γ = 1 + z − z , then min | z | = r Re p ( z ) ≥ min | z | = r Re q ( z ) . More related results are proved in [7, 15, 17]. For c > − f ∈ H [0 , F ∈ H [0 ,
1] given by Bernardi’s integral operator isdefined as(1.2) F ( z ) = c + 1 z c Z z t c − f ( t ) dt. Mathematics Subject Classification.
Key words and phrases.
Starlike functions, Briot–Bouquet differential subordination, Bernardi’s inte-gral operator, lemniscate of Bernoulli, parabolic starlike.
There is an important connection between Briot–Bouquet differential equations and theBernardi’s integral operator. If we set p ( z ) = zF ′ ( z ) /F ( z ) , where F is given by (1.2),then the functions f and p are related through the following Briot–Bouquet differentialequation zf ′ ( z ) f ( z ) = p ( z ) + zp ′ ( z ) p ( z ) + c . Several authors have investigated results on Briot–Bouquet differential subordination.For example, Ali et al. [3] determined the conditions on
A, B, D and E for p ( z ) ≺ (1 + Az ) / (1 + Bz ) when p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) is subordinate to (1 + Dz ) / (1 + Ez ),( A, B, D, E ∈ [ − , β so that p ( z ) is subordinate to e z or(1 + Az ) / (1 + Bz ) whenever 1 + βp ( z ) /p ′ ( z ) is subordinate to √ z or (1 + Az ) / (1 + Bz ) , ( − ≤ B < A ≤ γ and β so that p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) is subordinate to √ z implies p ( z ) ≺ e z . Conditions on A, B, β and γ are also determined so that p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) ≺ (1 + Az ) / (1 + Bz )implies p ( z ) ≺ √ z or e z . We determine conditions on A, B, β and γ so that p ( z ) ≺ (1 + Az ) / (1 + Bz ) , ( − ≤ B < A ≤
1) when p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) ≺ e z or ϕ P AR ( z ).The function ϕ P AR : D → C is given by(1.3) ϕ P AR ( z ) := 1 + 2 π (cid:18) log 1 + √ z − √ z (cid:19) , Im √ z ≥ ϕ P AR ( D ) = { w = u + iv : v < u − } = { w : Re w > | w − |} =: Ω P . As an appli-cation of our results, we give sufficient conditions for the Bernardi’s integral operator tobelong to the various subclasses of starlike functions which we define below.Let A be the class of all functions f ∈ H normalized by the conditions f (0) = 0 and f ′ (0) = 1. Let S denote the subclass of A consisting of univalent (one-to-one) functions.For an analytic function ϕ with ϕ (0) = 1, let S ∗ ( ϕ ) := (cid:26) f ∈ A : zf ′ ( z ) f ( z ) ≺ ϕ ( z ) (cid:27) . This class unifies various classes of starlike functions when Re ϕ >
0. Shanmugam [26]studied the convolution properties of this class when ϕ is convex while Ma and Minda[13] investigated the growth, distortion and coefficient estimates under less restrictiveassumption that ϕ is starlike and ϕ ( D ) is symmetric with respect to the real axis. Noticethat, for − ≤ B < A ≤
1, the class S ∗ [ A, B ] := S ∗ ((1 + Az ) / (1 + Bz )) is the classof Janowski starlike functions [9, 19]. For 0 ≤ α <
1, the class S ∗ [1 − α, −
1] =: S ∗ ( α )is the familiar class of starlike functions of order α , introduced by Robertson [22]. Theclass S ∗ := S ∗ (0) is the class of starlike function. The class S P := S ∗ ( ϕ P AR ) is the classof parabolic starlike functions, introduced by Rønning [25], consists of function f ∈ A satisfying Re (cid:18) zf ′ ( z ) f ( z ) (cid:19) > (cid:12)(cid:12)(cid:12)(cid:12) zf ′ ( z ) f ( z ) − (cid:12)(cid:12)(cid:12)(cid:12) , z ∈ D . Sok´ol and Stankiewicz [34] have introduced and studied the class S ∗ L := S ∗ ( √ z ); theclass S ∗ L consists of functions f ∈ A such that zf ′ ( z ) /f ( z ) lies in the region bounded RIOT–BOUQUET DIFFERENTIAL SUBORDINATION AND BERNARDI’S INTEGRAL OPERATOR 3 by the right-half of the lemniscate of Bernoulli given by Ω L := { w ∈ C : | w − | < } .Another class S ∗ e := S ∗ ( e z ), introduced recently by Mendiratta et al. [14], consists offunctions f ∈ A satisfying the condition | log( zf ′ ( z ) /f ( z )) | <
1. There has been severalworks [1, 2, 8, 20, 21, 27–33, 35] related to these classes.The following results are required in our investigation.
Lemma 1.1. [18, Theorem 2.1, p.2] Let Ω ⊂ C and suppose that ψ : C × D → C satisfiesthe condition ψ ( e e it , ke it e e it ; z ) / ∈ Ω , where z ∈ D , t ∈ [0 , π ] and k ≥ . If p ∈ H [1 , and ψ ( p ( z ) , zp ′ ( z ); z ) ∈ Ω for z ∈ D , then p ( z ) ≺ e z in D . Lemma 1.2. [23, Lemma 1.3, p.28] Let w be a meromorphic function in D , w (0) = 0 .If for some z ∈ D , max | z |≤| z | | w ( z ) | = | w ( z ) | , then it follows that z w ′ ( z ) /w ( z ) ≥ . Briot–Bouquet differential subordination
In the first result, we find conditions on the real numbers β and γ so that p ( z ) ≺ e z ,whenever p ( z ) + ( zp ′ ( z )) / ( βp ( z ) + γ ) ≺ √ z , where p ∈ H with p (0) = 1 . Thisresult gives the sufficient condition for f ∈ A to belong to the class S ∗ e by substituting p ( z ) = zf ′ ( z ) /f ( z ) . Theorem 2.1.
Let β, γ ∈ R satisfying max {− γ/e, − γe + e/ (1 − √ e ) } ≤ β ≤ − eγ . Let p ∈ H with p (0) = 1 . If the function p satisfies p ( z ) + zp ′ ( z ) βp ( z ) + γ ≺ √ z, then p ( z ) ≺ e z .Proof. Define the functions ψ : C × D → C and q : D → C as follows:(2.1) ψ ( r, s ; z ) = r + sβr + γ and q ( z ) = √ z so that Ω := q ( D ) = { w ∈ C : | w − | < } and ψ ( p ( z ) , zp ′ ( z ); z ) ∈ Ω for z ∈ D . Toprove p ( z ) ≺ e z , we use Lemma 1.1 so we need to show that ψ ( e e it , ke it e e it ; z ) / ∈ Ω whichis equivalent to show that | ( ψ ( e e it , ke it e e it ; z )) − | ≥
1, where z ∈ D , t ∈ [ − π, π ] and k ≥
1. A simple computation and (2.1) yield that ψ ( e e it , ke it e e it ; z ) = e e it + ke it e e it βe e it + γ ( − π ≤ t ≤ π )and(2.2) | ( ψ ( e e it , ke it e e it ; z )) − | =: f ( t ) g ( t ) ( − π ≤ t ≤ π ) , K. SHARMA, RASOUL AGHALARY, AND V. RAVICHANDRAN where f ( t ) = (cid:0) e t cos(2 sin t )(( γ + k cos t + βe cos t cos(sin t )) − ( k sin t + β sin(sin t ) e cos t ) ) − t ) e t ( k sin t + β sin(sin t ) e cos t )( γ + k cos t + βe cos t cos(sin t ))+ β sin (sin t ) e t − ( γ + βe cos t cos(sin t )) (cid:1) + (cid:0) e t cos(2 sin t )( k sin t + β sin(sin t ) e cos t )( γ + k cos t + βe cos t cos(sin t )) + sin(2 sin t ) e t (( γ + k cos t + βe cos t cos(sin t )) − ( k sin t + β sin(sin t ) e cos t ) ) − β sin(sin t ) e cos t ( γ + βe cos t cos(sin t )) (cid:1) and g ( t ) = ( β sin (sin t ) e t + ( γ + βe cos t cos(sin t )) ) . Define the function h : [ − π, π ] → R by h ( t ) = f ( t ) − g ( t ). Since h ( − t ) = h ( t ), we restrictto 0 ≤ t ≤ π . It can be easily verified that the function h attains its minimum valueeither at t = 0 or t = π . For k ≥
1, we have(2.3) h (0) = ( e ( eβ + γ + k ) − ( eβ + γ ) ) − ( eβ + γ ) and(2.4) h ( π ) = (cid:16)(cid:16) β/e + γ − ke (cid:17) − (cid:16) βe + γ (cid:17) (cid:17) − (cid:16) βe + γ (cid:17) . The given relation β ≥ − γ/e gives eβ + γ ≥ e ( k + eβ + γ ) > √ eβ + γ ) whichimplies e ( k + eβ + γ ) − ( eβ + γ ) > ( eβ + γ ) . Thus, the use of (2.3) yields h (0) > / (1 − √ e ) ≤ γ + β/e ≤ γ + β/e )(1 − √ e ) ≤ − k + γ + β/e ≤ − γ + β/e ≤ √ e ( γ + β/e ) which implies (( − k + γ + β/e ) /e ) ≥ γ + β/e ) which further implies (( − k + γ + β/e ) /e ) − ( γ + β/e ) ≥ ( γ + β/e ) . Hence,by using (2.4), we get that h ( π ) ≥ . So, h ( t ) ≥ , (0 ≤ t ≤ π ) and thus, (2.2) implies | ( ψ ( e e it , ke it e e it ; z )) − | ≥ p ( z ) ≺ e z .We will illustrate Theorem 2.1 by the following example: Example 2.2.
By taking β = 1 and γ = c ( c > − in Theorem 2.1, we get − /e +1 / (1 − √ e ) ≤ c ≤ − /e . By taking β = 1 , − /e + 1 / (1 − √ e ) ≤ γ ≤ − /e , n =1 , h ( z ) = √ z , a = 1 in [16, Theorem 3.2d, p.86], we get Re( aβ + γ ) > and βh ( z ) + γ ≺ R aβ + γ,n ( z ) , where R d,f ( z ) is the open door mapping given by R d,f ( z ) := d (1 + z ) / (1 − z ) + (2 f z ) / (1 − z ) . Thus by the use of [16, Theorem 3.2d, p.86], we get p ( z ) = − γ + Z t − γ e √ z +1 − √ tz +1 (cid:0) √ tz + 1 + 1 (cid:1) (cid:0) √ z + 1 + 1 (cid:1) dt which satisy the equation p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) = h ( z ) . Then p ( z ) ≺ e z . Suppose that the function F be given by Bernardi’s integral (1.2). Now we discussthe sufficient conditions for the function F to belong to various subclasses of starlikefunctions. We will illustrate the Theorem 2.1 by the following corollary. Corollary 2.3. (i) If the function f ∈ S ∗ L and the conditions of the Theorem 2.1 holdwith β = 1 and γ = c , then F ∈ S ∗ e . RIOT–BOUQUET DIFFERENTIAL SUBORDINATION AND BERNARDI’S INTEGRAL OPERATOR 5 (ii) If the function f ′ ( z ) ≺ √ z and the conditions of the Theorem 2.1 hold with β = 0 and γ = c + 1 , then F ′ ( z ) ≺ e z . Proof. ( i ) Let the function p : D → C be defined by p ( z ) = zF ′ ( z ) /F ( z ) . Then p isanalytic in D with p (0) = 1. Upon differentiating Bernardi’s integral given by (1.2), weobtain(2.5) ( c + 1) f ( z ) = zF ′ ( z ) + cF ( z ) . A computation now yields zf ′ ( z ) f ( z ) = p ( z ) + zp ′ ( z ) p ( z ) + c . By taking β = 1 and γ = c , the first part of the corollary follows from Theorem 2.1.( ii ) By defining a function p by p ( z ) = F ′ ( z ) and using (2.5), we get f ′ ( z ) = zF ′′ ( z ) c + 1 + F ′ ( z ) . By taking β = 0 and γ = c + 1, the result follows from Theorem 2.1.In the following result, we derive conditions on the real numbers A, B , β and γ so that p ( z ) + ( zp ′ ( z )) / ( βp ( z ) + γ ) ≺ e z implies p ( z ) ≺ (1 + Az ) / (1 + Bz ) , ( − ≤ B < A ≤ , where p ∈ H with p (0) = 1 . This result gives the sufficient condition for f ∈ A to belongto the class S ∗ [ A, B ] by substituting p ( z ) = zf ′ ( z ) /f ( z ) . Theorem 2.4.
Let − < B < A ≤ and β, γ ∈ R . Suppose that(i) (cid:0) A − B (cid:1) / (cid:0) (1 ∓ B )((1 ∓ A ) β + (1 ∓ B ) γ ) (cid:1) ≥ ± (1 ∓ A ) / (1 ∓ B ) + e. (ii) β (1 ± A ) + γ (1 ± B ) > .Let p ∈ H with p (0) = 1 . If the function p satisfies p ( z ) + zp ′ ( z ) βp ( z ) + γ ≺ e z , then p ( z ) ≺ (1 + Az ) / (1 + Bz ) .Proof. Define the functions P and w as follows:(2.6) P ( z ) = p ( z ) + zp ′ ( z ) βp ( z ) + γ and w ( z ) = p ( z ) − A − Bp ( z )so that p ( z ) = (1 + Aw ( z )) / (1 + Bw ( z )) . Clearly, w ( z ) is analytic in D with w (0) = 0.In order to prove p ( z ) ≺ (1 + Az ) / (1 + Bz ), we need to show that | w ( z ) | < D . Ifpossible, suppose that there exists z ∈ D such thatmax | z |≤| z | | w ( z ) | = | w ( z ) | = 1 , then by Lemma 1.2, it follows that there exists k ≥ z w ′ ( z ) = kw ( z ) . Let w ( z ) = e it , ( − π ≤ t ≤ π ) and G := Aβ + Bγ . A simple calculation and by using (2.6),we get(2.7) P ( z ) = ke it ( A − B ) + (1 + Ae it ) ( β + γ + Ge it )(1 + Be it ) ( β + γ + Ge it ) =: u + iv ( − π ≤ t ≤ π ) . K. SHARMA, RASOUL AGHALARY, AND V. RAVICHANDRAN
We derive a contradiction by showing | log P ( z ) | ≥
1. This later inequality is equivalentto(2.8) f ( t ) := 4(arg( u + iv )) + (log (cid:0) u + v (cid:1) ) − ≥ − π ≤ t ≤ π ) . From (2.7), we get u = 1( B + 2 B cos t + 1) (( β + γ ) + G + 2 G ( β + γ ) cos t ) (cid:0) G ( A + B )( β + γ ) cos 2 t + cos t (cid:0) A ( BG (2( β + γ ) + k ) + G + ( β + γ )( β + γ + k )) − B Gk + 2 G ( β + γ )+ B (cid:0) G − ( β + γ )( − β − γ + k ) (cid:1) (cid:1) + ( β + γ ) (cid:0) AB ( β + γ + k ) + β − B k + γ (cid:1) + G ( AB + 1) + G ( A ( β + γ + k ) + B ( β + γ − k )) (cid:1) and v = ( A − B ) sin t ( − BGk + G + 2 G ( β + γ ) cos t + ( β + γ )( β + γ + k ))( B + 2 B cos t + 1) (( β + γ ) + G + 2 G ( β + γ ) cos t ) . Substituting these values of u and v in (2.8), we observe that f ( t ) is an even function of t and so, it is enough to show that f ( t ) ≥ t ∈ [0 , π ]. It can be easily verified thatthe function f ( t ) attains its minimum value either at t = 0 or t = π . We show that both f (0) and f ( π ) are non negative. Note that, for k ≥ f (0) = − ψ ( k )) + (log( ψ ( k ))) and(2.10) f ( π ) = − − φ ( k ))) + (log( φ ( k ))) , where ψ ( k ) := (cid:0) A β + A (2 β + Bγ + γ + k )+ β + B ( γ − k )+ γ (cid:1) / (cid:0) (1+ B )( β (1+ A )+ γ (1+ B )) (cid:1) and φ ( k ) := (cid:0) A β − Aβ + ( A − B − γ − Ak + β + Bk (cid:1) / (cid:0) ( B − − Aβ + β − Bγ + γ ) (cid:1) .The function ψ is increasing as ψ ′ ( k ) = ( A − B ) / (cid:0) (1 + B )( β (1 + A ) + γ (1 + B )) (cid:1) > ii ) and therefore, the given hypothesis ( i ) yields that ψ ( k ) ≥ ψ (1) = (1 + A ) / (1 + B ) + ( A − B ) / (cid:0) (1 + B )( β (1 + A ) + γ (1 + B )) (cid:1) ≥ e which gives thatarg ψ ( k ) = 0 and (log( ψ ( k ))) ≥ (2 log e ) = 4 . Thus, the use of (2.9) yields f (0) ≥ . The function φ is increasing as φ ′ ( k ) = ( A − B ) / (cid:0) (1 − B )( β (1 − A ) + γ (1 − B )) (cid:1) > ii ) and therefore, the given hypothesis ( i ) yields that φ ( k ) ≥ φ (1) = − (1 − A ) / (1 − B ) + ( A − B ) / (cid:0) (1 − B )( β (1 − A ) + γ (1 − B )) (cid:1) ≥ e which furtherimplies arg( − φ ( k )) = π and (log( φ ( k ))) ≥ (2 log e ) = 4 . Hence, by using (2.10), we get f ( π ) ≥ π > . This completes the proof.We will illustrate Theorem 2.4 by the following example:
Example 2.5.
By taking A = 1 / , B = − / , β = 1 and γ = c ( c > − in Theorem 2.4,we get − / ≤ c ≤ (1 − e ) / (1 + 3 e ) . By taking β = 1 , − / ≤ γ ≤ (1 − e ) / (1 + 3 e ) , n = 1 , h ( z ) = e z , a = 1 in [16, Theorem 3.2d, p.86], we get Re( aβ + γ ) > and βh ( z ) + γ ≺ R aβ + γ,n ( z ) , where R d,f ( z ) is the open door mapping given by R d,f ( z ) := d (1 + z ) / (1 − z ) + (2 f z ) / (1 − z ) . Thus by the use of [16, Theorem 3.2d, p.86], we get p ( z ) = Z t − γ e − Chi ( tz )+ Chi ( z ) − Shi ( tz )+ Shi ( z ) dt − γ RIOT–BOUQUET DIFFERENTIAL SUBORDINATION AND BERNARDI’S INTEGRAL OPERATOR 7 which satisy the equation p ( z ) + zp ′ ( z ) / ( βp ( z ) + γ ) = h ( z ) . Then p ( z ) ≺ (2 + z ) / (2 − z ) .Here, Chi ( z ) and Shi ( z ) are the hyperbolic cosine integral function and the hyperbolicsine integral function respectively defined as follows: Chi ( z ) = η + log( z ) + Z z cosh( t ) − t dt and Shi ( z ) = Z z sinh( t ) t dt, where η is the Euler’s constant. The next corollary is obtained by substituting p ( z ) = zf ′ ( z ) /f ( z ) with γ = 0, B = 0and A = 1 − α , (0 ≤ α <
1) in Theorem 2.4.
Corollary 2.6.
Let ≤ α < and β > satisfy the conditions α + e + β − ≤ ( αβ ) − and − α ≥ β (2 − α )( e − α ) . If the function f ∈ A satisfies the subordination zf ′ ( z ) f ( z ) + 1 β (cid:18) zf ′′ ( z ) f ′ ( z ) − zf ′ ( z ) f ( z ) (cid:19) ≺ e z , then f ∈ S ∗ α . Our next corollary deals with the class R [ A, B ] defined by R [ A, B ] = (cid:26) f ∈ A : f ′ ( z ) ≺ Az Bz (cid:27) . The two parts of the following corollary are obtained by taking p ( z ) = zF ′ ( z ) /F ( z ) with β = 1, γ = c and p ( z ) = F ′ ( z ) with β = 0, γ = c + 1 respectively in Theorem 2.4. Corollary 2.7. (i) If the function f ∈ S ∗ e and the conditions of the Theorem 2.4 holdwith β = 1 and γ = c , then F ∈ S ∗ [ A, B ] . (ii) The function f ′ ( z ) ≺ e z and the conditions of the Theorem 2.4 hold with β = 0 and γ = c + 1 , then F ∈ R [ A, B ] . In the next result, we find the conditions on the real numbers
A, B , β and γ so that p ( z ) ≺ √ z , whenever p ( z ) + ( zp ′ ( z )) / ( βp ( z ) + γ ) ≺ (1 + Az ) / (1 + Bz ), − ≤ B < A ≤
1, where p ∈ H with p (0) = 1 . As an application of the next result, it provides sufficientconditions for f ∈ A to belong to the class S ∗ L . Theorem 2.8.
Let − ≤ B < A ≤ and β, γ ∈ R satisfy the following conditions:(i) √ − β − √ − γ ≥ B ( − A (2 β + √ γ ) + B (1 + 4( √ β + γ ))) .(ii) (1 + 4( √ − β − √ − γ ) ≥ ( − A (2 β + √ γ ) + B (1 + 4( √ β + γ ))) . Let p ∈ H with p (0) = 1 . If the function p satisfies p ( z ) + zp ′ ( z ) βp ( z ) + γ ≺ Az Bz , then p ( z ) ≺ √ z .Proof. Define the functions P and w as follows:(2.11) P ( z ) = p ( z ) + zp ′ ( z ) βp ( z ) + γ and w ( z ) = p ( z ) − K. SHARMA, RASOUL AGHALARY, AND V. RAVICHANDRAN which implies p ( z ) = p w ( z ) . Clearly, w ( z ) is analytic in D with w (0) = 0. In orderto complete our proof, we need to show that | w ( z ) | < D . Assume that there exists z ∈ D such that max | z |≤| z | | w ( z ) | = | w ( z ) | = 1 , then by Lemma 1.2, it follows that there exists k ≥ z w ′ ( z ) = kw ( z ) . Let w ( z ) = e it , ( − π ≤ t ≤ π ). By using (2.11), we get P ( z ) = p w ( z ) + zw ′ ( z )2 p w ( z )( β p w ( z ) + γ ) . A simple computation shows that P ( z ) = ke it + 2 (1 + e it ) (cid:0) γ + β √ e it (cid:1) √ e it (cid:0) γ + β √ e it (cid:1) ( − π ≤ t ≤ π )and(2.12) (cid:12)(cid:12)(cid:12)(cid:12) P ( z ) − A − BP ( z ) (cid:12)(cid:12)(cid:12)(cid:12) =: f ( t ) g ( t ) ( − π ≤ t ≤ π ) , where f ( t ) = (cid:0) (2 β cos t + 2( β − γ )) sin(arg(1 + e it ) / p t/ t ( k + 2( γ + β ( − e it ) / p t/ (cid:1) + (cid:0) − cos t ( k + 2( γ + β ( − e it ) / p t/ β sin t sin(arg(1 + e it ) / p t/
2) + 2( β − γ )(1 − cos(arg(1 + e it ) / p t/ (cid:1) and g ( t ) = (cid:0) − A ( β sin t + γ sin(arg(1 + e it ) / p t/ Bβ cos ( t/ p t/ e it ) /
2) + B sin t ( k + 2 γ + 2 β cos(arg(1 + e it ) / p t/ (cid:1) + (cid:0) − Aβ cos ( t/
2) + B ( k + 2 γ ) cos t + 2 γ − Bβ sin t sin(arg(1 + e it ) / p t/
2) + 2( − Aγ + Bβ cos t + β ) cos(arg(1 + e it ) / p t/ (cid:1) . Define h ( t ) = f ( t ) − g ( t ). Since h ( t ) is an even function of t , we restrict to 0 ≤ t ≤ π .It can be easily verified that for both the cases ( i ) and ( ii ), the function h ( t ) attains itsminimum value either at t = 0 or t = π . Note that for k ≥ h ( π ) = (1 − B ) k > h (0) = (4( √ − β − √ − γ + k ) − ( B (4( √ β + γ ) + k ) − A (2 β + √ γ )) =: S ( k ) . The function S ′ is increasing as S ′′ ( k ) = 2(1 − B ) > i ) yields that S ′ ( k ) ≥ S ′ (1) = 2(1 + 4( √ − β − √ − γ ) − B ( − A (2 β + √ γ ) + B (1 + 4( √ β + γ ))) ≥ S ( k ) ≥ S (1) = (1 + 4( √ − β − √ − γ ) − ( − A (2 β + √ γ ) + B (1 + 4( √ β + γ ))) . Thus, the use of given condition ( ii )and (2.13) yields h (0) ≥ . So, h ( t ) ≥ t ∈ [0 , π ] and therefore, (2.12) implies | ( P ( z ) − / ( A − BP ( z )) | ≥
1. This contradicts the fact that P ( z ) ≺ (1 + Az ) / (1 + Bz )and completes the proof. RIOT–BOUQUET DIFFERENTIAL SUBORDINATION AND BERNARDI’S INTEGRAL OPERATOR 9
The next corollary is obtained by substituting p ( z ) = zf ′ ( z ) /f ( z ) with γ = 0, A =1 − α , (0 ≤ α <
1) and B = − Corollary 2.9.
Let f ∈ A . If the function f satisfies the subordination zf ′ ( z ) f ( z ) + 1 β (cid:18) zf ′′ ( z ) f ′ ( z ) − zf ′ ( z ) f ( z ) (cid:19) ≺ − α ) z − z (cid:18) α − √ ≤ β < , ≤ α < (cid:19) , then f ∈ S ∗ L . By taking p ( z ) = zF ′ ( z ) /F ( z ) with β = 1 and γ = c in Theorem 2.8 gives the followingcorollary: Corollary 2.10.
Let − ≤ B < A ≤ satisfy the following conditions:(i) √ − − √ − c ≥ B ( − A (2 + √ c ) + B (1 + 4( √ c ))) .(ii) (1 + 4( √ − − √ − c ) ≥ ( − A (2 + √ c ) + B (1 + 4( √ c ))) . If f ∈ S ∗ [ A, B ] then F ∈ S ∗ L . By taking p ( z ) = F ′ ( z ) with β = 0 and γ = c + 1 in Theorem 2.8 gives the followingcorollary: Corollary 2.11.
Suppose that − ≤ B < A ≤ satisfy the following conditions:(i) − √ − √ − c ≥ B ( − √ c + 1) A + (5 + 4 c ) B ) .(ii) (5 − √ − √ − c ) ≥ ( − √ c + 1) A + (5 + 4 c ) B ) . If f ∈ R [ A, B ] then F ′ ( z ) ≺ √ z . In the next result, we compute the conditions on the real numbers
A, B , β and γ so that p ( z ) + ( zp ′ ( z )) / ( βp ( z ) + γ ) ≺ (1 + Az ) / (1 + Bz ) , ( − ≤ B < A ≤
1) implies p ( z ) ≺ e z ,where p ∈ H with p (0) = 1 . As an application of the next result, it provides sufficientconditions for f ∈ A to belong to the class S ∗ e . Theorem 2.12.
Let − ≤ B < A ≤ and β, γ ∈ R satisfy the following conditions:(i) e β (1 − B ) + e ( − B ( − Aβ + Bγ + B ) − β + γ + 1) + γ ( AB − ≥ .(ii) ( e (( A + e − β − ( eβ + 1) B + 1) + γ ( A + e (1 − B ) − e ( − ( A − e + 1) β + B ( eβ +1) + 1) + γ ( − A + e ( B + 1) − ≥ . (iii) e ( β (1 − AB ) + B ( γ − − γ + 1) + e γ (1 − AB ) + β ( B − ≥ . (iv) ( e (( A − β + (1 − B )( γ − e ( A − γ + β (1 − B ))( − e (( A + 1) β + ( B + 1)(1 − γ )) − e ( A + 1) γ + β ( B + 1)) ≥ .Let p ∈ H with p (0) = 1 . If the function p satisfies p ( z ) + zp ′ ( z ) βp ( z ) + γ ≺ Az Bz , then p ( z ) ≺ e z .Proof. Define the functions ψ : C × D → C and q : D → C as follows:(2.14) ψ ( r, s ; z ) = r + sβr + γ and q ( z ) = 1 + Az Bz so that Ω := q ( D ) = { w ∈ C : | ( w − / ( A − Bw ) | < } and ψ ( p ( z ) , zp ′ ( z ); z ) ∈ Ω for z ∈ D . To prove p ( z ) ≺ e z , we use Lemma 1.1 so we need to show that ψ ( e e it , ke it e e it ; z ) / ∈ Ω which is equivalent to show that | ( ψ ( e e it , ke it e e it ; z ) − / ( A − Bψ ( e e it , ke it e e it ; z )) | ≥ z ∈ D , t ∈ [ − π, π ] and k ≥
1. A simple computation and (2.14) yield that ψ ( e e it , ke it e e it ; z ) = e e it + ke it e e it βe e it + γ ( − π ≤ t ≤ π )and(2.15) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ψ ( e e it , ke it e e it ; z ) − A − Bψ ( e e it , ke it e e it ; z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) =: f ( t ) g ( t ) ( − π ≤ t ≤ π ) , where f ( t ) = e t (2 βk sin t sin(sin t ) + 2 βk cos t cos(sin t ) − β cos(sin t ) + 2 βγ cos(sin t ))+ e t (( β − γ ) + k − βk cos t + 2 γk cos t + 2 βγ sin (sin t ) − βγ cos (sin t ))+ e cos t (2 γk sin t sin(sin t ) − γk cos t cos(sin t ) + 2 βγ cos(sin t ) − γ cos(sin t ))+ β e t + γ and g ( t ) = A γ + β B e t + 2 βBe t (( Bγ − Aβ ) cos(sin t ) + Bk cos( t − sin t ))+ e t ( B ( B ( γ + k ) − Aβγ ) + 2 B ( Bγ − Aβ ) k cos t − ABβγ cos(2 sin t )+ A β ) + 2 Aγe cos t (( Aβ − Bγ ) cos(sin t ) − Bk cos( t + sin t )) . Define h ( t ) = f ( t ) − g ( t ). Since h ( − t ) = h ( t ), we restrict to 0 ≤ t ≤ π . It can be easilyverified that the function h ( t ) attains its minimum value either at t = 0 or t = π . For k ≥
1, we have h (0) = e ((1 − A ) β + 2 k ( β ( AB −
1) + (1 − B ) γ ) + 4 βγ ( AB −
1) + (1 − B )( γ + k ))+ 2 eγ ( − A β + ( AB − γ + k ) + β ) + 2 e β ( β ( AB −
1) + (1 − B )( γ + k ))+ e β (1 − B ) + (1 − A ) γ =: φ ( k )(2.16)and h ( π ) = − e ( e (( A − β + (1 − B )( γ − k )) + e ( A − γ + β (1 − B ))( e ((1 + A ) β + ( B + 1)( k − γ )) + e ( A + 1) γ − β ( B + 1)) =: ψ ( k ) . (2.17)The function φ ′ is increasing as φ ′′ ( k ) = 2(1 − B ) e > i ) yields that φ ′ ( k ) ≥ φ ′ (1) = 2 e ( e ( − B ( − Aβ + Bγ + B ) − β + γ + 1) + γ ( AB −
1) + e β (1 − B )) ≥ φ ( k ) ≥ φ (1) = ( e (( A + e − β − ( eβ + 1) B + 1) + γ ( A + e (1 − B ) − e ( − ( A − e + 1) β + B ( eβ + 1) + 1) + γ ( − A + e ( B + 1) − ii ) and (2.16) yields h (0) ≥ . In view of ( iii ), observe that ψ ′′ ( k ) = 2(1 − B ) /e > ψ ′ ( k ) = ψ ′ (1) = 2( e ( β (1 − AB ) + B ( γ − − γ + 1) + e γ (1 − AB ) + β ( B − /e ≥ ψ ( k ) = ψ (1) = (( e (( A − β +(1 − B )( γ − e ( A − γ + β (1 − B ))( − e (( A +1) β + ( B + 1)(1 − γ )) − e ( A + 1) γ + β ( B + 1))) /e . Hence, the use of given condition( iv ) and (2.17) yields that h ( π ) ≥ . So, h ( t ) ≥ , (0 ≤ t ≤ π ) and thus, (2.15) implies | ( ψ ( e e it , ke it e e it ; z ) − / ( A − Bψ ( e e it , ke it e e it ; z )) | ≥ p ( z ) ≺ e z . RIOT–BOUQUET DIFFERENTIAL SUBORDINATION AND BERNARDI’S INTEGRAL OPERATOR11
The next corollary is obtained by substituting p ( z ) = zf ′ ( z ) /f ( z ) with γ = 0, B = 0and A = 1 − α , (0 ≤ α <
1) in Theorem 2.12.
Corollary 2.13.
Suppose ≤ α < and β ≥ / (1 − e ) satisfy the conditions ( − αβ + βe + 1)( β ( α + e −
2) + 1) ≥ and ( β − e ((2 − α ) β + 1))( β + e ( − αβ − ≥ . If thefunction f ∈ A satisfies the condition (cid:12)(cid:12)(cid:12)(cid:12) zf ′ ( z ) f ( z ) + 1 β (cid:18) zf ′′ ( z ) f ′ ( z ) − zf ′ ( z ) f ( z ) (cid:19) − (cid:12)(cid:12)(cid:12)(cid:12) < − α, then f ∈ S ∗ e . The two parts of the following corollary are obtained by taking p ( z ) = zF ′ ( z ) /F ( z )with β = 1, γ = c and p ( z ) = F ′ ( z ) with β = 0, γ = c + 1 respectively in Theorem 2.12. Corollary 2.14. (i) If the function f ∈ S ∗ [ A, B ] and the conditions of the Theo-rem 2.12 hold with β = 1 and γ = c , then F ∈ S ∗ e . (ii) The function f ∈ R [ A, B ] and the conditions of the Theorem 2.12 hold with β = 0 and γ = c + 1 , then F ′ ( z ) ≺ e z . In the next result, we find the conditions on the real numbers
A, B , β and γ so that p ( z ) ≺ (1 + Az ) / (1 + Bz ) , ( − ≤ B < A ≤ p ( z ) + ( zp ′ ( z )) / ( βp ( z ) + γ ) ∈ Ω P ,where p ∈ H with p (0) = 1 . As an application of the next result, it provides sufficientconditions for f ∈ A to belong to the class S ∗ [ A, B ]. Theorem 2.15.
Let − ≤ B < A ≤ and β, γ ∈ R . For k ≥ and ≤ m ≤ , assumethat G := Aβ + Bγ, L := k + β + γ . Further assume that(i) BG ( β + γ ) > . (ii) (cid:0) G ( A L + 4( β + γ )) − B ( AGL + 2( β + γ ) + 2 G ) + B G (4( β + γ ) + L ) (cid:1)(cid:0) G (cid:0) A L − β + γ ) (cid:1) + B (cid:0) − AGL + 4( β + γ ) + 4 G (cid:1) + B G ( L − β + γ )) (cid:1) ≥ G ( A − B ) (cid:0) GL ( A L − β + γ )) − B ( AGL + 2 G ( L − β + γ )) − L ( β + γ )( − β − γ + 2 L )) + B GL ( L − β + γ )) (cid:1) . (iii) G ( A − B ) ( β + γ + k ) ≤ B − G ( β + γ ) + 2 B ( β + γ − G ) . (iv) β + γ ≥ , G ≥ . (v) m ( A − B ) ( β + γ + G + 1) ≥ ( B + 1) ( β + γ + G ) . Let p ∈ H with p (0) = 1 . If the function p satisfies p ( z ) + zp ′ ( z ) βp ( z ) + γ ≺ ϕ P AR ( z ) , then p ( z ) ≺ (1 + Az ) / (1 + Bz ) .Proof. Define the functions P and w as given by the equation (2.6) which implies p ( z ) =(1 + Aw ( z )) / (1 + Bw ( z )) . Proceeding as in Theorem 2.4, we need to show that | w ( z ) | < D . If possible suppose that there exists z ∈ D such thatmax | z |≤| z | | w ( z ) | = | w ( z ) | = 1 , then by Lemma 1.2, it follows that there exists k ≥ z w ′ ( z ) = kw ( z ) . Let w ( z ) = e it , ( − π ≤ t ≤ π ). A simple calculation and by using (2.6), we get(2.18) P ( z ) = ke it ( A − B ) + (1 + Ae it ) ( β + γ + Ge it )(1 + Be it ) ( β + γ + Ge it ) ( − π ≤ t ≤ π ) . Define the function h by(2.19) h ( z ) = u + iv = p ( P ( z ) − π / . We show that | ( e h ( z ) − / ( e h ( z ) + 1) | ≥
1; this condition is same as the inequalityRe e h ( z ) ≤
0. This last inequality is indeed equivalent to cos v ≤ / ≤ | v/π | ≤ h given in (2.19) together with (2.18), we get(2.20) | v | π = √ A − B | m ( t ) || Ge it + L | / √ | Be it | / | Ge it + β + γ | / ( − π ≤ t ≤ π ) , where m ( t ) = sin (arg (( e it ( A − B )( Ge it + L )) / ((1 + Be it )( Ge it + β + γ ))) / . (a) We will first show that | v/π | ≤ | m ( t ) | ≤ f ( t ) ≥ − π ≤ t ≤ π ), where f ( t ) = 4(1 + B + 2 B cos t )(( β + γ ) + G + 2( β + γ ) G cos t ) − ( A − B ) ( L + G + 2 LG cos t ) . After substituting x = cos t ( − π ≤ t ≤ π ), the above inequality reduces to F ( x ) ≥ x with − ≤ x ≤
1, where F ( x ) = 4(1 + B + 2 Bx )(( β + γ ) + G + 2( β + γ ) Gx ) − ( A − B ) ( L + G + 2 LGx ) . A simple computation shows that for x = G ( A L − β + γ )) − B ( AGL + 2( β + γ ) + 2 G ) + B G ( L − β + γ ))16 BG ( β + γ ) ,F ′ ( x ) = 0 and F ′′ ( x ) = 32 BG ( β + γ ) > i ). Therefore, F ( x ) ≥ F ( x ). Observe that F ( x ) = 116 BG ( β + γ ) (cid:0)(cid:0) G (cid:0) A L + 4( β + γ ) (cid:1) − B (cid:0) AGL + 2( β + γ ) + 2 G (cid:1) + B G (4( β + γ ) + L ) (cid:1)(cid:0) G (cid:0) A L − β + γ ) (cid:1) + B G ( L − β + γ ))+ B (cid:0) − AGL + 4( β + γ ) + 4 G (cid:1) (cid:1) − G ( A − B ) (cid:0) GL (cid:0) A L − β + γ ) (cid:1) + B GL ( L − β + γ )) − B (cid:0) AGL + 2 G ( L − β + γ )) − L ( β + γ )( − β − γ + 2 L ) (cid:1) (cid:1)(cid:1) and F ( x ) ≥ ii ).(b) We will next show that | v/π | ≥ / g ( t ) ≥ − π ≤ t ≤ π ), where g ( t ) = 4( A − B ) m ( t )( L + G +2 LG cos t ) − (1+ B +2 B cos t )(( β + γ ) + G +2( β + γ ) G cos t )After substituting x = cos t ( − π ≤ t ≤ π ) and m = m ( t ), the above inequality reduces to H ( x ) ≥ x with − ≤ x ≤
1, where H ( x ) = 4( A − B ) m ( L + G + 2 LGx ) − (1 + B + 2 Bx )(( β + γ ) + G + 2( β + γ ) Gx ) . RIOT–BOUQUET DIFFERENTIAL SUBORDINATION AND BERNARDI’S INTEGRAL OPERATOR13
In view of ( i ), ( iii ), ( iv ) and the fact that − ≤ m ≤
1, we see that H ′′ ( x ) = − BG ( β + γ ) < H ′ ( x ) ≤ H ′ ( −
1) = 8 m G ( A − B ) ( β + γ + k ) − B − G ( β + γ ) − B ( − G + β + γ ) ≤
0. Thus, H ( x ) ≥ H (1) = 4 m ( A − B ) ( β + γ + G + k ) − ( B + 1) ( β + γ + G ) =: ψ ( k ). Using ( iv ), we observe that ψ ′′ ( k ) = 8 m ( A − B ) ≥ k ≥
1, we have ψ ′ ( k ) ≥ ψ ′ (1) = 8 m ( A − B ) ( β + γ + G + 1) ≥
0. Thus by using ( v ), weget H ( x ) ≥ ψ ( k ) ≥ ψ (1) = 4 m ( A − B ) ( β + γ + G + 1) − ( B + 1) ( β + γ + G ) ≥ p ( z ) = zf ′ ( z ) /f ( z ) with γ = 0, B = − A = 1 − α , (0 ≤ α <
1) in Theorem 2.15.
Corollary 2.16.
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Email address : [email protected]; [email protected] Department of Mathematics, Faculty of Science, Urmia University, Urmia, Iran
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