# Extension of Nunokawa Lemma for Functions with Fixed Second Coefficient and Its Applications

Authors: Mostafa Amani; Rasoul Aghalary; Ali Ebadian
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study some properties of analytic functions with fixed initial coefficients. The methodology of differential subordination is used for modification and improvements of several well-known results for subclasses of univalent functions by restricting the functions with fixed initial coefficients. Actually, by extending the Nunokawa lemma for fixed initial coefficient functions, we obtain some novel results on subclasses of univalent functions, such as differential inequalities for univalency or starlikeness of analytic functions. Also, we provide some new sufficient conditions for strongly starlike functions. The results of this paper extend and improve the previously known results by considering functions with fixed second coefficients.


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## Body

## 1. Introduction and Preliminaries

LetH be the class of analytic functions in the unit disc U=z:z<1. For a∈ℂ and n∈ℕ, let us define two well-known classes of analytic functions as follows:(1)Ha,n=f∈H:fz=a+anzn+⋯,z∈U,An=f∈H:fz=z+an+1zn+1,z∈U.We denote byA=A1 and S⊂A, as the class of univalent functions. Also, we denote by S∗α, the set of starlike function of order α (α<1), as follows:(2)S∗α=f∈A:Rezf′zfz>α,z∈U,which is introduced by Robertson in [1]. Let S∗0=S∗ be the classes of starlike functions. Brannan and Kirwan in [2] introduced SS∗λ the class of strongly starlike function of order λ, 0<λ<1, by(3)SS∗λ=f∈A:argzf′zfz<λ,z∈U.Also, Takahashi and Nunokawa in [3] defined the following subclasses of A:(4)SS∗α,λ=f∈A:−απ2<argzf′zfz<λπ2,z∈U.It is easy to see thatSS∗minα,λ⊂SS∗α,λ⊂SS∗maxα,λ. Also, if α=λ, then SS∗α,λ⊂S∗⊂S. Then, they are the subclasses of univalent functions in the unit disc U.Recently, Ali et al. [4] have extended the theory of second-order differential subordination for functions with a fixed initial coefficient. We denote by Hβa,n the class of analytic functions with a fixed initial coefficient as follows:(5)Hβa,n=p∈H:pz=a+βzn+an+1zn+1+⋯.Further, let(6)An,b=f∈H:fz=z+bzn+1+⋯,z∈U,and also, let(7)Sn,β∗α,λ=f∈SS∗α,λ:fz=z+βzn+1+⋯,z∈U,where b∈ℂ is fixed. We denote by Ab=A1,b, and we assume that β and b are positive real numbers.The importance of the second coefficient of analytic functions was shown in the monograph [5], for example, coefficient estimates in two well-known growth and distortion theorems for functions in the class S. On the other hand, one of the significant tools in geometric function theory is the theory of differential subordination due to Miller and Mocanu [6]. Recently, Ali et al. [4] improved the theory of differential subordination by this assumption that the second coefficient of analytic function is fixed. For some applications of this improvement, see [7, 8]. In addition, Nunokawa has proved a theorem known as Nunokawa lemmas [9] which have had many applications in the geometric function theory. So, it is natural to ask if it is possible to extend this lemma for analytic functions with a fixed second coefficient.In this paper, we extend this lemma, and then, it will be applied to obtain several new results by restricting the functions to have a fixed second coefficient. It is a remarkable fact when we restrict ourselves to the functions with a fixed second coefficient; then, it will expand the validity domain of the results. As a byproduct of this idea, one can also get results for other class mapping. Even though, in this paper, we are not going into details of all such applications, we remark that other applications will appear in other papers of the authors.In [10], Silverman examined the class Gb, consisting of normalized functions f∈S that satisfy the condition(8)1+zf″z/f′zzf′z/fz−1<bz∈U,for some positive b. He has proved that Gb⊆S∗2/1+1+8b.For a fixed coefficientβ, let(9)Gbn,β=f∈Gb:fz=z+βzn+1+⋯,z∈U.Forfz and gz in H, we say that fz is subordinate to gz if there exists a function wz∈H with w0=0,wz<1z∈U, such that fz=gwzz∈U denote by fz≺gz. When gz is univalent in U, then we can see(10)fz≺gz⇔f0=g0,fU⊆gU.This article is arranged as follows. In Section2, we extend the lemma due to Nunokawa [9] for fixed second coefficient functions and then will bring its nice applications for finding sufficient conditions for strongly starlike functions. In Section 3, by considering the special cases of this extension, we will provide some sufficient conditions for starlikeness and univalency of analytic functions. Also, we discuss about preserving starlikeness of the general integral operator Iδ,γ. Finally, in Theorem 26, we state a new result which is improvement of some well-known results in the literature.We just need a definition and a fundamental lemma due to Ali et al. [4] to get the results.Definition 1 ([6], Definition 1, p. 158).
LetQ be the class of functions q that are analytic and injective in U¯\Eq, where(11)Eq≔ζ∈∂U:limz⟶ζqz=∞,and are such that q′ζ≠0 for ζ∈∂U\Eq.Lemma 2 (see [4]).
Letq∈Q with q0=a, and p∈Hca,n with pz≡a. If there exist a point z0∈U such that pz0∈q∂U and pz:z<z0⊂qU, then(12)z0p′z0=mζ0q′ζ0,Re1+z0p″z0p′z0≥mRe1+ζ0q″ζ0q′ζ0,where q−1pz0=ζ0=eiθ0 (θ is real) and(13)m≥n+q′0−cz0nq′0+cz0n.


## 2. Main Results

The Nunokawa lemma is a very useful tool in the theory of differential subordination that was proved for the first time by Nunokawa [9]. At first, we mention extension of this lemma with its proof because we need it in the next theorems.Theorem 3.
Letp∈Hβ1,n and pz≠0 in U. If there exist z1∈U and z2∈U such that z1=z2=r and for z∈z:z<r(14)−πα2=argpz1<argpz<argpz2=πλ2,with 0<α≤1,0<λ≤1, and β is a real number with 0≤β≤λ+αsinπα/α+λ. Then, we have(15)z1p′z1pz1=−iλ+α2m1,z2p′z2pz2=iλ+α2m2,where(16)m1≥n+sinπα/λ+α−β/λ+αsinπα/λ+α+β/λ+αtanπλ2λ+α,m2≥n+sinπα/λ+α−β/λ+αsinπα/λ+α+β/λ+αtanπα2λ+α.Proof.
Let us define(17)qz=1−e−2πiα/λ+αz1−zλ+α/2,with 0<α≤1,0<λ≤1. It is easy to check that q is analytic in U with q0=1 and(18)−πα2<argqU<πλ2.
Also,q∈Q with Eq=1. According to properties of q and assumption (14), we have pz1∈q∂U, pz2∈q∂U, and pz:∣z∣<∣z0∣⊂qU. Set(19)p1z=exp−iπλ−α2λ+αpz2/λ+α,q1z=exp−iπλ−α2λ+α1−e−2πiα/λ+αz1−z.
Then, it can be verified thatq1U is the right half plane Rew>0, p1z1=−ix1 with x1>0, p2z2=ix2 with x2>0 and p1z:z<z0⊂q1U. We notice that(20)exp−iπλ−α2λ+α=−iexpiπαλ+α=iexpiπλλ+α,and so, by considering c=exppiπα/λ+α, we can rewrite the functions p1 and q1 as(21)p1z=−icpz2/λ+α,q1z=−ic−c¯z1−z.
Now, taking the derivative ofq1 and calculating the inverse of q1 yields(22)q1′z=2Imc1−z2,q1−1z=iz−ciz−c¯.
Sincep1∈Hc1a,n with(23)c1=2exp−iπα−λ/2α+λα+λβ,a=exp−iπα−λ2α+λ,and so, from Lemma 2, we deduce that there exist two complex numbers ζ1 and ζ2 on ∂U with p1z1=q1ζ1 and p1z2=q1ζ2 such that(24)z1p1′z1=k1ζ1q1′ζ1,z2p1′z2=k2ζ2q1′ζ2,where(25)k1≥n+q1′0−c1z1nq1′0+c1z1n,k2≥n+q1′0−c1z2nq1′0+c1z2n.
But(26)ζ1=q1−1p1z1=x1−cx1−c¯;hence,(27)z1p′z1pz1=λ+α2z1p1′z1p1z1=k1λ+α2ζ1q1′ζ1p1z1=k1λ+α2x1−cx1−c¯×1−ix1×2Imc1−x1−c/x1−c¯2=−ik1α+λ2x12+2x1cosπλ/α+λ+12x1sinπλ/α+λ.
But the function(28)fx=x12+2x1cosπλ/α+λ+12x1sinπλ/α+λtakes its minimum at the point x=1, and we have f1=tanπλ/2α+λ. In view of q1′0=2sinαπ/α+λ and (25), we obtain(29)m1=k1fx>k1f1=n+sinπα/λ+α−β/λ+αsinπα/λ+α+β/λ+αtanπλ2λ+α;thus, we showed that(30)z1p′z1pz1=−iλ+α2m1,m1≥n+sinπα/λ+α−β/λ+αsinπα/λ+α+β/λ+αtanπλ2λ+α.
In a similar way, by noting thatpz2=ix2 with x2>0 and letting(31)ζ2=q−1ix2=x2+cx2+c¯,we can see that(32)z2p′z2pz2=λ+α2×z2p1′z2p1z2=k2λ+α2ζ2q1′ζ2p1z2=k2λ+α2×x2+cx2+c¯×1ix2×2Imc1−x2+c/x2+c¯2=ik2α+λ2x22+2x2cosπα/α+λ+12x2sinπα/α+λ.
Now, doing analogous above, we find that(33)m2≥n+sinπα/λ+α−β/λ+αsinπα/λ+α+β/λ+αtanπα2λ+α.
Thus, we showed that(34)z2p′z2pz2=−iλ+α2m2,m2≥n+sinπα/λ+α−β/λ+αsinπα/λ+α+β/λ+αtanπα2λ+α,which is (16); therefore, the proof is complete.☐Remark 4.
Nunokawa [9] provided a similar result to Theorem 3 with this assumption that p∈H1,n. But, if we fix the second initial coefficient, then we may have some reform as seen above.Theorem 5.
Let0<λ≤1, 0<α≤1and0≤β≤λ+α/nsinπα/α+λ. If f∈Gbn,β with(35)b=δminm1a~11−δ+2a~1−δcosπλ/2δ+a~1−1−δ2sinπλ/2δ,m2a~21−δ+2a~2−δcosπα/2δ+a~2−1−δ2sinπα/2δ,where δ=λ+α/2 and(36)m1=n+sinπα/2δ−nβ/2δsinπα/2δ+nβ/2δtanπλ4δ,m2=n+sinπα/2δ−nβ/2δsinπα/2δ+nβ/2δtanπα4δ,a~1=1−δ2sin2πα/2δ−δcosπα/2δ1−δ,a~2=1−δ2sin2πλ/2δ−δcosπλ/2δ1−δ,then f∈SSn,β∗α,λ.Proof.
Suppose thatf∈Gbn,β. Let us define(37)pz=zf′zfz,p1z=cpz2/λ+α,with(38)c=exp−iπλ−α2λ+α.
Then, it is clear thatp∈Hnβ1,n,p1∈Hc1a,n with(39)c1=2exp−iπα−λ/2α+λα+λnβ,a=exp−iπα−λ2α+λ.
Also, it is easy to see that(40)zp′zp2z=1+zf″z/f′zzf′z/fz−1.
Iff∉SSn,β∗α,λ, then p1U is not contained in the right half plane Rew>0; hence, there exists a point z1∈U such that p1z:z<z1 is contained in the right half plane Rew>0 while, p1z1=−ix1, or p1z1=ix2, where x1,x2>0. In the first step, let p1z1=−ix1, then by applying the same argument as Theorem 3, we obtain(41)z1p′z1pz1=λ+α2z1p1′z1p1z1=k1α+λ2x12+2x1cosπλ/α+λ+12x1sinπλ/α+λ,where(42)k1≥m1=n+sinπα/2δ−nβ/2δsinπα/2δ+nβ/2δtanπλ4δ.
Also, from the relation ofp and p1, we have ∣pz1∣=x1δ, and so, by making use of (41), it yields(43)z1p′z1p2z1=k1α+λ2x11−δ+2x1−δcosπλ/α+λ+x1−1−δ2sinπλ/α+λ.
If we define(44)gx=x11−δ+2x1−δcosπλα+λ+x1−1−δ,then one can verify that g takes its minimum at a~1; hence, applying (42) and (43), we have(45)z1p′z1p2z1≥δm1a~11−δ+2a~1−δcosπλ/2δ+a~1−1−δ2sinπλ/2δ≥b,which contradicts with the assumption f∈Gbn,β.
Ifp1z0=ix2, where x2>0, using a similar argument as above leads the contradiction. Indeed, applying the previous considerations, we obtain(46)z1p′z1pz1=λ+α2z1p1′z1p1z1=k2α+λ2x12+2x1cosπα/α+λ+12x1sinπα/α+λ,where(47)k2≥m2=n+sinπα/2δ−nβ/2δsinπα/2δ+nβ/2δtanπα4δ.
Notice thatpz1=x2δ, and so, by making use of (46), it yields(48)z1p′z1p2z1=k2α+λ2x21−δ+2x2−δcosπα/α+λ+x2−1−δ2sinπα/α+λ.
If we define(49)hx=x21−δ+2x2−δcosπαα+λ+x2−1−δ,then by some calculation, we find that h takes its minimum at a~2; hence, applying (47) and (48), we have(50)z1p′z1p2z1≥δm2a~21−δ+2a~2−δcosπα/2δ+a~1−1−δ2sinπα/2δ≥b,which contradicts with the assumption f∈Gbn,β. So, in both cases, we come to a contradiction, and the proof is complete.☐Lettingα=λ in the Theorem 5, we get the following corollary.Corollary 6.
Let0<α≤1 and 0≤β≤2α/n. If f∈Gbn,β with(51)b=n+2α−nβ2α+nβα1−α1−α1+α1+α,then f∈SSn,β∗α.Puttingα=β=1/2 and n=1 in Corollary 6, we obtain the following result.Corollary 7.
Iff∈A1,1/2, and(52)1+zf″z/f′zzf′z/fz−1<4349,then f is strongly starlike of order 1/2.In the following theorem, we mention a strong result which provides sufficient conditions to strongly starlike functions.Theorem 8.
Let0<λ<1, 0<α<1 and 0≤β≤λ+α/nsinπα/α+λ. If f∈An,β and(53)1+zf″z/f′zzf′z/fz−1∈gUz∈U,where(54)gz=b2γz1−z21−δ,γ=expiπα−λ2−λ−α,and b,δ are defined as Theorem 5. Then,(55)f∈SSn,β∗α,λ.Proof.
Since(56)gz=b2γz1−z21−δ,we have(57)geiθ=biγsinθ1−δ.
Now, for0<θ<π, we have(58)geiθ=tbexpiπ21−δ=ω1t,with t≥1, and for π<θ<2π,(59)geiθ=tbexp−iπ21−α=ω2t,with t≥1. Hence, gU is a whole complex plane except the half lines ω1t and ω2t. Suppose that(60)pz=zf′zfz,p1z=cpz2/λ+α,with(61)c=expiπλ−α2λ+α.
Then, it is clear thatp∈Hnβ1,n,p1∈Hc1a,n with(62)c1=2exp−iπα−λ/2α+λα+λnβ,a=exp−iπα−λ2α+λ.
Also, it is easy to see that(63)zp′zp2z=1+zf″z/f′zzf′z/fz−1.
Iff∉SSn,β∗α,λ, then p1U is not contained in the right half plane Rew>0; hence, there exists a point z1∈U such that p1z:z<z1 is contained in the right half plane Rew>0 while p1z1=−ix1, or p1z1=ix2, where x1,x2>0. In the first step, let p1z1=−ix1; then, using the same argument as Theorem 3, we obtain(64)z1p′z1pz1=λ+α2z1p1′z1p1z1=−ik1α+λ2x12+2x1cosπλ/α+λ+12x1sinπλ/α+λ,where(65)k1≥m1=n+sinπα/2δ−nβ/2δsinπα/2δ+nβ/2δtanπλ4δ.
Also, from the relation ofp and p1, we have(66)pz1=x1δexp−π2α,and so, by making use of (64), it yields(67)argz1p′z1p2z1=exp−π21−α,z1p′z1p2z1=k1α+λ2x11−δ+2x1−δcosπλ/α+λ+x1−1−δ2sinπλ/α+λ≥b,which contradicts with the assumption.
For the casep1z1=ix2, where x2>0, using the same argument as Theorem 3, we obtain(68)z1p′z1pz1=iλ+α2×z1p1′z1p1z1=ik2α+λ2x12+2x1cosπα/α+λ+12x1sinπα/α+λ,where(69)k2≥m2=n+sinπα/2δ−nβ/2δsinπα/2δ+nβ/2δtanπα4δ.
Notice that(70)pz1=x2δexpiπ2λ,and so, by making use of (68), we have(71)argz1p′z1p2z1=expiπ21−λ,z1p′z1p2z1=k2α+λ2x21−δ+2x2−δcosπα/α+λ+x2−1−δ2sinπα/α+λ≥b,which contradicts with the assumption. So, in both cases, we come to a contradiction, and the proof is complete.☐


## 3. Starlikeness of Analytic Functions

There are many differential inequalities in geometric function theory which is used for univalency or starlikeness of analytic functions. For example, Ozaki [11] has proved that if f∈A satisfies the condition Re1+zf″z/f′z>−1/2, then f is univalent. Also, the well-known result known as the Mark-strohhäcker [5] states that f∈A, then(72)Re1+zf″zf′z>0⟹Rezf′zfz>12.Hence, it is natural to extend the similar results to analytic functions with a fixed initial coefficient.In this section, we try to obtain some inequalities in analytic functions with second fixed coefficients which improve earlier results obtained in the literature. At first, we bring the following corollary.Corollary 9.
Letp∈Hβ1,n and z0∈U such that Repz0>0forz<z0 and Repz0=0 with pz0≠0. Then, we have(73)z0p′z0pz0=im,m≥n+2−β2+β,(74)z0p′z0≤−12n+2−β2+β1+pz02.Proof.
By lettingα=λ=1 in Theorem 3, we obtain (73). For (74), we see that pz0 is purely imaginary so pz0¯=−pz0 and pz02=−pz02. Let us put(75)qz=1+z1−z.
Then, we have thatq∈Q with q′0=2. From the assumption and Lemma 2, there exist ζ0∈∂U with pz0=qζ0 such that(76)z0p′z0=mζ0q′ζ0=mq−1pz021−q−1pz02=−12m1+pz02,which shows that(77)−z0p′z0≥12n+2−β2+β1+pz02,and z0p′z0 is a negative real number and (74) is obtained.☐Remark 10.
The special case whenβ=2 reduces to the result due to Nunokawa [12].Corollary 11.
Letp∈Hβ1,n and pz≠0 in U. If there exist z0∈U such that argpz<πα/2 for z<z0 and argpz0=πα/2, where α>0. Then, we have(78)z0p′z0pz0=−iαm,m≤−12a+1an+2α−β2α+β,when argpz0=−πα/2 and(79)z0p′z0pz0=iαm,m≥12a+1an+2α−β2α+β,when argpz0=πα/2, where pz01/α=±ia.Proof.
Letqz=pz1/α. We have Reqz>0 for z<z0 and Reqz0=0. Let Reqz0=±ia with a>0. From Corollary 9, we have(80)z0q′z0qz0=1αz0p′z0pz0=im⟹z0p′z0pz0=iαm,where(81)m≤−12a+1an+2α−β2α+β,whenqz0=−ia,m≥12a+1an+2α−β2α+β,whenqz0=+ia.
So, the proof is done.☐Remark 12.
Note the special case of Corollary11 when β=2α reduces to the well-known lemma [9].Theorem 13.
Letp∈Hβ1,n,0≤β<2, and suppose that(82)Imzp′zpz<n+2−β2+βz∈U.
Then, we haveRepz>0.Proof.
From the assumption (82), we have pz≠0. In fact, if pz has a zero of order n at z=α, then we have(83)pz=z−αnp1z,where p1z is analytic in U and p1α≠0. Then, we have(84)zp′zpz=nzz−α+zp1′zp1z.
Butn is fixed, so we have(85)limz⟶αImnzz−α=∞.
Then, the imaginary part of the right-hand side of (84) can take any values when z approaches α. This contradicts (82). This shows that pz≠0 for z∈U. Therefore, if there exists a point z0∈U such that Repz0=0, then we have pz0≠0. From Corollary 9, we obtain(86)z0p′z0pz0=im,m≥n+2−β2+β.
This contradicts (82). This completes our proof.☐Remark 14.
The special case whenβ=2 reduces to the result obtained in [9].Corollary 15.
Letf∈An,b,0≤b<2, and suppose that(87)Imzf′zfz<n+2−b2+bz∈U.
Then, we haveRefz/z>0 for z∈U.Proof.
Letf∈An,b and put pz=fz/z. Then, pz∈Hb1,n. Applying Theorem 13, we get the result.☐Corollary 16.
Letf∈An,b,0≤b<4/n+1 and suppose that(88)Imzf″zf′z<2n+24−bn+14+bn+1z∈U.
Then, we haveRef′z>0 for z∈U.Proof.
Letf∈An,b and put pz=f′z. Then, pz∈H1/2n+1b1,n. Applying Theorem 13, we get the result.☐Corollary 17.
Letf∈An,b,0≤b<2/n, and suppose that(89)Im1+zf″zf′z−zf′zfz<n+2−nb2+nbz∈U.
Then, we haveRezf′z/fz>0 for z∈U.Proof.
Letf∈An,b and put pz=zf′z/fz. Then, pz∈Hnb1,n. Applying Theorem 13, we get the result.☐Theorem 18.
Letp∈Hβ1,n and pz≠0 in U. Also, suppose that 0≤γ<δ<γ+2 and 0≤β≤21−α. If there exist z0∈U such that Repz>α for z<z0 and Repz0=α but pz0≠α where 0<α<1. Then, we have(90)Rez0p′z0δpz0+γ≤Γα=−δα+γm21−αδ2,α∈0,δ−γ2,α−1m2δα+γ,α∈δ−γ2,1,where(91)m≥n+2−2α−β2−2α+β.Proof.
Let(92)qz=1−2α−1z1−z,where 0<α<1. We know that q is analytic and univalent in U with q′0=2−2α and ReqU>α. So, q∈Q with Eq=1. From Lemma 2, there exist ζ0 on ∂U with pz0=qζ0 such that(93)z0p′z0=mζ0q′ζ0,wherem≥n+q′0−βz0nq′0+βz0n.
We have(94)z0p′z0=mζ0q′ζ0=m×pz0−1pz0−2α−1×2−2α1−pz0−1/pz0−2α−12=m2−2αpz0−1pz0−2α−1,where(95)m≥n+2−2α−β2−2α+β.
By takingpz0=α+it for a fixed real t, we have(96)Rez0p′z0δpz0+γ=−δα+γm2−2αt2+α−12δ2t2+δα+γ2,m≥n+2−2α−β2−2α+β.
IfA=t2+α−12/δ2t2+δα+γ2, then for α∈0,δ−γ/2 and for all real number t, we have A≥1/δ2 which leads to(97)Rez0p′z0δpz0+γ≤−δα+γm21−αδ2,0<α≤δ−γ2,m≥n+2−2α−β2−2α+β.
Also, we haveα−1/δα+γ2≤A<1 when α∈δ−γ/2,1, and t is an arbitrary real number. This leads to(98)Rez0p′z0δpz0+γ≤α−1m2δα+γ,δ−γ2<α<1,m≥n+2−2α−β2−2α+β.
Therefore, the proof is complete.☐Corollary 19.
Letp∈Hβ1,n and pz≠0inU. Suppose also that 0<λ,0≤γ<δ<γ+2, 0<α<1,0≤β≤21−α and(99)Repz+λzp′zδpz+γ>α+λ−δα+γm12δ21−α,α∈0,δ−γ2,α+λα−1m12δα+γ,α∈δ−γ2,1,where(100)m1=n+2−2α−β/2−2α+β.
Then, we haveRepz>α for ∣z∣<1.Proof.
If there exists a point,z0, ∣z0∣<1, such that(101)Repz>αz<z0,Repz0=α.
Then, from Theorem18, we have(102)Rez0p′z0δpz0+γ≤−δα+γm121−αδ2,α∈0,δ−γ2,α−1m12δα+γ,α∈δ−γ2,1,and so,(103)Repz0+λz0p′z0δpz0+γ≤α+λ−δα+γm12δ21−α,α∈0,δ−γ2,α+λα−1m12δα+γ,α∈δ−γ2,1.
This contradicts (99), and the proof is complete.☐By takingδ=1,γ=0, and pz=zf′z/fz in Corollary 19, we obtain the following theorem.Theorem 20.
Letf∈An,b, 0<α≤1/2 and 0≤b<21−α/n. If(104)Re1−λzf′zfz+λ1+zf″zf′z>α+λ−αm121−α,where(105)m1=n+2−2α−nb2−2α+nb,0<λ.
Then,(106)Rezf′zfz>α.Remark 21.
We remark that the special case of Theorem20 with α=1/2,λ=1, and n=1 was proved in [8]. Also, the special case when α=1/2,λ=n=1, and b=0 reduces to the well-known Mark-strohhäcker result [13].Also, by takingδ=1,γ=0, and pz=zf′z/fz in Corollary 19, we obtain the following theorem.Theorem 22.
Letf∈An,b, 1/2<α<1, and 0≤b<21−α/n. If(107)Re1−λzf′zfz+λ1+zf″zf′z>α+λα−1m12α,where(108)m1=n+2−2α−nb2−2α+nb,0<λ.
Then,(109)Rezf′zfz>α.
Forγ≥0,δ>0, let Iδ,γ be the integral operator defined as(110)Fz=Iδ,γfz=δ+γzγ∫0zfδttγ−1dt1/δ.
We note that the integral operatorIδ,γ maps An,b to An,δ+γb/n+δ+γ. Now, we state the next following theorems.Theorem 23.
Letf∈An,b, 0<α≤1/2, and 0≤b<2n+δ+γ1−α/nδ+γ. If0≤γ<δ<γ+2 and(111)Rezf′zfz>α+−δα+γm12δ21−α,where(112)m1=n+2−2α−nbδ+γ/n+δ+γ2−2α+nbδ+γ/n+δ+γ.
Then,(113)RezF′zFz>α,where Fz=Iδ,γfz is defined in (110).Proof.
Let us definepz=zF′z/Fz. It is easy to show that(114)pz+zp′zδpz+γ=zf′zfz.
Now considering Corollary19, we get our result.☐Puttingδ=b=n=1, α=1/2, and γ=0 in the above theorem, we have the following result.Corollary 24.
Letf∈A1,1. If(115)Rezf′zfz>−16.
Then,(116)RezF′zFz>12.
By considering Corollary19, we have the following result.Theorem 25.
Letf∈An,b, 1/2<α≤1, and0≤b<2n+δ+γ1−α/nδ+γ. If0≤γ<δ<γ+2 and(117)Rezf′zfz>α+λα−1m12δα+γ,where(118)m1=n+2−2α−nbδ+γ/n+δ+γ2−2α+nbδ+γ/n+δ+γ.
Then,(119)RezF′zFz>α,where Fz=Iδ,γfz is defined in (110).Finally, we state the following theorem which is very interesting and in the especial case improves some earlier results in the literature.Theorem 26.
Suppose thatf∈An,b and 0≤b<1/1+n. Also, let(120)Re1+zf″zf′z<m1+22,where(121)m1=n+1−b1+n1+b1+n.
Ifzf′z/fz is analytic in U and omits 21+m1/2+m1, then f is starlike in the unit disc U and(122)Rezf′zfz<21+m12+m1.Proof.
Let us define the functionp as(123)zf′zfz=21+m1pzm1+2+m1pz.
Then, it is easy to see thatp∈H2b1+n1,n, and(124)1+zf″zf′z=zf′zfz+zp′zpz−2+m1zp′zm1+2+m1pz.
If there exists a pointz0 such that Repz>0 for z<z0 and Repz0=0. Then, using Corollary 9, we have(125)z0p′z0pz0=ik,where k is a real number with k≥m1. For the case argpz0=π/2, and pz0=ia with a>0, we obtain(126)1+z0f″z0f′z0=z0f′z0fz0+z0p′z0pz0−2+m1z0p′z0m1+2+m1pz0=21+m1iam1+2+m1ia+z0p′z0pz0−z0p′z0pz02+m1pz0m1+2+m1pz0=21+m1iam1−2+m1iam12+2+m12a2+ik+k2+m1am1−2+m1iam12+2+m12a2.
Therefore, we obtain(127)Re1+z0f″z0f′z0≥21+m12+m1a2m12+2+m12a2+m122+m11+a22m12+2+m12a2=2+m12,and that contradicts the hypothesis of the theorem. In the case argpz0=−π/2 and pz0=−ia with a>0, by applying the same method as above, we also obtain(128)Re1+z0f″z0f′z0≥2+m12,and this also contradicts with the assumption of the theorem. Hence, we obtain(129)Repz>0z∈U.
But the inequality(130)zf′zfz−1+m12+m1<1+m12+m1holds true if only if(131)21+m1pzm1+2+m1pz−1+m12+m1<1+m12+m1,or equivalently(132)2+m1pz−m1m1+2+m1pz<1.
Finally, we note that the last inequality holds true if and only ifRepz>0, and this is proved in the above. Therefore, from (130), we have(133)0<zf′zfz<21+m12+m1,and the proof is complete.☐Puttingn=1,b=0 in Theorem 26, we obtain the following result.Corollary 27.
Suppose thatf∈A1,0 and(134)Re1+zf″zf′z<2.
Ifzf′z/fz is analytic in U and omits 4/3, then f is starlike in the unit disc U and(135)Rezf′zfz<43.Remark 28.
Note that Corollary27 improves the result of Sumit and Ravichandran [8].



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*Source: 4564694-2021-07-27.xml*