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Research Article

Majorization problem for general family of functions with bounded radius rotations

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Pages 388-392 | Received 23 Jan 2024, Accepted 29 Jun 2024, Published online: 13 Jul 2024

Abstract

In this article, we study the majorization problem for the general class of functions with bounded radius rotation, which the authors introduced here. Furthermore, the coefficient bound for majorized functions associated with this class is derived. Consequently, we present new results as corollaries and point out relevant connections between the main results obtained from the ones in the literature.

MSC Classification:

1. Introduction and preliminaries

Let A denote the class of all analytic functions f in the open unit disc :={tC:|t|<1} satisfying the normalized condition f(0)=0=f(0)1 whose series representation is given as: (1) f(t)=t+n=2antn,t.(1)

Let SA be the class of all univalent functions in . A function f is subordinate to a function gA if there exists an analytic function ω: with ω(0)=0, such that f(t)=g(t)°ω(t),t. In addition, if g is univalent in , then f(t)g(t) if and only if f(0)=g(0) and f()g(). Let f have the form Equation(1) and Sf(t) be defined by Sf(t)=tf(t)/f(t). The class S of all starlike functions in was one of the earliest subfamilies of S. Functions fS are characterized analytically with the condition Re(Sf(t))>0 in . This class was first comprehensively introduced and studied by Shanmugam (Citation1989) using the properties of convolution and subordination, and later, coefficients-related results of this general family were established by Ma and Minda (Citation1992). For this general class, they considered analytic function ϕ with the following properties:

  1. ϕS with Reϕ(t)>0 in ;

  2. ϕ is starlike with respect to ϕ(0)=1;

  3. ϕ() is symmetric about the real axis, and

  4. ϕ(0)>0.

Hence, they introduced the class (2) S(ϕ)={fA:Sf(t)ϕ(t),t}.(2)

Consider the function ϕj:C defined by ϕ1(t)=et,ϕ2(t)=(1+st)2,s(0,1/2],ϕ3(t)=t+1+t2 and  ϕ4(t)=(1+At)/(1+Bt),1B<A1, which reduces to ϕ5(t)=(1+t)/(1t) for A=1,B=1 and ϕ6(t)=(1+(12α)t)/(1t),α[0,1) for A=12α,B=1. For ϕ=ϕj (j=1,2,,6), the class S(ϕ) becomes the classes Se,SL(s),SC,S(A,B),S and Sα, respectively. The geometric properties of these families were discussed by Janowski (Citation1973), Masih and Kanas (Citation2020), Mendiratta, Nagpal, and Ravichandran (Citation2015), Raina and Sokół (Citation2015), and Robertson (Citation1936).

A natural extension of the class S is the class Sκ of functions with bounded radius rotaion. Geometrically speaking, fSκ means that the total variation of angle between radius vector f(reiθ) makes with positive real axis is bounded by κπ. Also, fSκ if and only if for κ2, it has a representation: Sf(t)=(κ+24)p1(t)(κ24)p2(t), where pj(t)ϕ5(t)(j=1,2) (Pinchuk, Citation1971). Recently, this class received attention in the direction of Ma and Minda’s class. Afis and Noor (Citation2020) and Jabeen and Saliu (Citation2022) introduced the classes SCκ and SLκ(s) of bounded radius rotation related to crescent and limaçon domains. They investigated inclusion properties, radius problems, coefficient inequalities, and other associated geometrical properties. For more information on the study of Sκ and its subfamilies, one may see “A survey on functions of bounded boundary and bounded radius rotation” by Noor and Malik§ (Citation2012).

Let f,gA. We say that f is majorized by g or g majorized f (written as f(t)g(t)) in if |f(t)||g(t)| for all t. The study of this concept started with the work of MacGregor (Citation1967), where he showed that if f(t)g(t) in , then there exists an analytic function φ with |φ(t)|1 in such that f(t)=φ(t)g(t),t.

This article opened the door to study the majorization problems for various classes of univalent functions (see Adegani, Alimohammadi, Bulboacă, & Cho, Citation2022; Cho, Oroujy, Analouei Adegani, & Ebadian, Citation2020; Gangania & Kumar, Citation2022; Gupta, Nagpal, & Ravichandran, Citation2021; Hameed Mohammed & Analouei Adegani, Citation2023). Cho et al. (Citation2020) examined the majorization property for the class S(ϕ), which was later extended by Hameed Mohammed and Analouei Adegani (Citation2023) and Adegani et al. (Citation2022).

Motivated by the recent work (Afis & Noor, Citation2020; Cho et al., Citation2020; Jabeen & Saliu, Citation2022), we introduce and investigate the majorization problem for the general class Sκ(ϕ) of functions of bounded radius rotation. Furthermore, we obtain the coefficient inequality for majorized functions associated with this class and illustrate some particular cases of our results.

Definition 1.1.

Let fA. Then fSκ(ϕ) if and only for κ2, Sf(t)=(κ+24)p1(t)(κ24)p2(t), where pj(t)ϕ(t)(j=1,2).

As a special case, we have the following:

  1. For ϕ(t)=ϕ2(t), we have the class introduced and studied by Jabeen and Saliu (Citation2022).

  2. For ϕ(t)=ϕ3(t), we have the class introduced and studied by Afis and Noor (Citation2020).

  3. For ϕ(t)=ϕ4(t), we obtain the class introduced and studied by Noor (Citation2009).

  4. For κ=2, we arrive at the general family of starlike functions introduced and investigated by Ma and Minda (Citation1992).

Remark 1.

When we replace ϕ(t) with Ma and Minda functions, we obtain the corresponding subclasses of bounded radius rotation of Ma and Minda type.

Lemma 1.2.

[Kanas & Sugawa, Citation2006, Proposition 5.4] If ϕ is convex univalent, has non-negative Taylor coefficients about the origin and Re(ϕ(t))>0 in , then min|t|=r|ϕ(t)|=ϕ(r)|ϕ(t)|ϕ(r)=max |t|=r|ϕ(t)|.

2. Main results

Theorem 2.1.

Let fA be majorized by gSκ(ϕ). Then g(t) majorized f(t) in the disc |t|<rκ, where rκ is the smallest positive root of the equation: (3) (1r2)|(κ+2)min|t|=r|ϕ(t)|(κ2)max|t|=r|ϕ(t)||8r=0.(3)

Proof.

Since f(t)g(t), then f(t)=φ(t)g(t) for some analytic function φ with |φ(t)|1,t. Therefore, (4) f(t)=φ(t)g(t)+φ(t)g(t)=(φ(t)g(t)g(t)+φ(t))g(t).(4)

But gSκ(ϕ) implies that (5) Sg(t)=(κ+24)p1(t)(κ24)p2(t),pj(t)ϕ(t)(j=1,2)=(κ+24)ϕ(ω1(t))(κ24)ϕ(ω2(t)),(5) where ωj (j = 1, 2) is analytic in with ωj(0)=0 such that |ωj(t)|=r<1. Since Re(ϕ(ωj(t)))>0 (j=1,2) in , we can rewrite Equation(5) as (6) g(t)g(t)=t(κ+24)ϕ(ω1(t))(κ24)ϕ(ω2(t)).(6)

From the minimum and maximum modulus principle, we have that min|t|=r|ϕ(t)|min|ω1(t)|=r|ϕ(ω1(t))|=min|ω1(t)|r|ϕ(ω1(t))| and max|ω2(t)|r|ϕ(ω2(t))|=max|ω2(t)|=r|ϕ(ω2(t))|max |t|=r|ϕ(t)|.

In view of these inequalities, Equation(6) becomes (7) |g(t)g(t)|r|(κ+24)|ϕ(ω1(t))|(κ24)|ϕ(ω2(t))||4r|(κ+2)min|ω1(t)|=r|ϕ(ω1(t))|(κ2)max|ω2(t)|=r|ϕ(ω2(t))||4r|(κ+2)min|t|=r|ϕ(t)|(κ2)max|t|=r|ϕ(t)||.(7)

From Equation(4) and Equation(7), we obtain |f(t)|(4r|φ(t)||(κ+2)min|t|=r|ϕ(t)|(κ2)max|t|=r|ϕ(t)||+|φ(t)|)|g(t)|.

Let |φ(t)|=ρ<1, and using the Schwarz Pick lemma [Nehari, Citation1952, p. 168], we arrive at |f(t)|(4r(1ρ2)(1r2)|(κ+2)min|t|=r|ϕ(t)|(κ2)max|t|=r|ϕ(t)||+ρ)|g(t)|.

Let V(r,ρ)=4r(1ρ2)(1r2)|(κ+2)min|t|=r|ϕ(t)|(κ2)max|t|=r|ϕ(t)||+ρ,r(0,1),ρ[0,1].

To find rκ, we choose rκ=max{r(0,1):V(r,ρ)1,ρ[0,1]}. But V(r,ρ)1 if and only if 0(1r2)|(κ+2)min|t|=r|ϕ(t)|(κ2)max|t|=r|ϕ(t)||4r(1+ρ):=Wκ(r,ρ).

It is easy to see that Wκ(r,ρ) is decreasing on 0ρ1. Thus,

minWκ(r,ρ)=Wκ(r,1):=Wκ(r), where Wκ(r)=(1r2)|(κ+2)min|t|=r|ϕ(t)|(κ2)max|t|=r|ϕ(t)||8r.

We note that Wκ(0)>0 and Wκ(1)<0. Therefore, there exists r=rκ which satisfies Equation(3). Hence, Wκ(r)0 for all r in the disc |t|<rκ.

It is worthy of note that for κ=2, Theorem 2.1 reduces to Theorem 2 in Cho et al. (Citation2020) and also becomes various corollaries obtained therein for respective Ma and Minda functions. Now for κ>2, we obtain the following results.

Corollary 2.2.

Let fA and f(t)g(t) with gSκ(ϕ6):=Sκ(α). Then f(t)g(t) in the disc |t|<rκ(α), where rκ(α)={2k(1α)+2+(1α)2κ2+4(1α)κ+8αα122κ+2,α=12 or rκ(α)={2k(1α)2+(1α)2κ24(1α)κ+8αα12and0ακ4κ2withκ>42κ+2,α=12

Proof.

Since ϕ6(t) is convex in , then by Lemma 1.2 and Theorem 2.1, we have the required result. □

For α=0 in Corollary 2.2, we arrive at the following result.

Corollary 2.3.

Let fA and f(t)g(t) with gSκ(ϕ5):=Sκ. Then f(t)g(t) in the disc |t|<rκ, where rκ=2k+2+κ(κ+4) or rκ=2k2+κ(κ4)withκ>4.

Corollary 2.4.

Let fA and f(t)g(t) with gSκ(ϕ2):=SLκ(s). Then g(t) majorized f(t) in the disc |t|<rκ(s), where rκ(s) is the smallest roots of the equation 1(κs+2)r(1s2)r2+ksr3s2r4=0 or 1+(κs+2)r+(1s2)r2ksr3+s2r4=0.

Proof.

The method of proof is the same as the proof of Corollary 2.2. □

Corollary 2.5.

Let fA and f(t)g(t) with gSκ(ϕ1):=Seκ. Then g(t) majorized f(t) in the disc |t|<rκ(s), where rκ is the smallest roots of the equation (1r2)|4cosh(r)κsinh(r)|8r=0.

Theorem 2.6.

Let ϕ(t)=1+B1t+B2t2+B3t3+ be convex in and fA have the form Equation(1). If f(t)g(t) with gSκ(ϕ), then (8) |an|1+m=2n[j=2m(2(j2)+κ|B1|2(n1)!)].(8)

Proof.

Let (9) Sg(t)=p(t)=1+n=1dntn.(9)

Since gSκ(ϕ), then there exist pj(t)=1+n=1cn,jtnϕ(t),j=1,2 such that (10) 1+n=1dntn=(κ+24)(1+n=1cn,1tn)(κ24)(1+n=1cn,2tn).(10)

Comparing the coefficients of tn in Equation(10), we arrive at (11) |dn|(κ+24)|cn,1|+(κ24)|cn,2|.(11)

Since pj(t)ϕ(t),j=1,2 and ϕ(t) is convex, then by Rogosinski’s lemma (Rogosinski, Citation1945), we have that |cn,j||B1|,j=1,2. Consequently Equation(11) becomes |dn|κ2|B1|.

Let g(t)=z+n=2bntn. Then Equation(9) implies z+n=2nbntn=(z+n=2bntn)(1+n=1dntn).

Now, following the procedures of Kuroki and Owa (Citation2012), we find that (12) |bn|j=2n(2(j2)+κ|B1|2(n1)!).(12)

Let φ(t)=n=0entn be analytic in with|φ(t)|1 in . Then the condition f(t)g(t) in implies that f(t)=φ(t)g(t), which follows that (13) an=e0bn+e1bn1+e2bn2++en2b2+en1.(13)

Let Γ be any circle in |t|=r,(0<r<1). Then by Cauchy derivative formula, we can write Equation(13) as |an|=|12πiΓφ(t)tn(1+b2t+b3t2+b4t3++bntn1) dt|12π02π|φ(reiθ)|rn1|1+b2t+b3t2+b4t3++bntn1| dθ1rn1(1+m=2n|bm|).

Since this inequality holds for all r(0,1), then by Equation(12), it follows that |an|1+m=2n|bm|1+m=2n[j=2m(2(j2)+κ|B1|2(n1)!)], which conclude the required result. □

Remark 2.

For κ=2, Theorem 2.6 becomes Theorem 3 of Cho et al. (Citation2020).

For κ>2, we have the following results:

Corollary 2.7.

Let fA and suppose f(t)g(t) in with gSκ(ϕ4):=Sκ(A,B). Then |an|1+m=2n[j=2m(2(j2)+κ(AB)2(n1)!)].

Proof.

It is easy to see that ϕ4(t)=1+(AB)n=1(1)n+1Bn1tn.

Since ϕ4 is a function that maps onto a circular domain and so, it is a convex function. Therefore, by Rogosinski’s lemma (Rogosinski, Citation1945), B1=AB. Thus, in view of Theorem 2.6, we have the required result. □

Corollary 2.8.

Let fA and suppose f(t)g(t) in22:={zC:|t|<2/2} with gSκ(ϕ3):=SCκ. Then |an|1+m=2n[j=2m(2(j2)+κ2(n1)!)].

Proof.

Raina and Sokół (Citation2019) showed that the function ϕ3 is convex in the disc 22. Thus, by Rogosinski’s lemma (Rogosinski, Citation1945), B1=1. Consequently, from Theorem 2.6, we obtain the desired result. □

Corollary 2.9.

Let fA and suppose f(t)g(t) in with gSL@@κ. Then for 0<s1/2, |an|1+m=2n[j=2m((j2)+κs(n1)!)].

Proof.

We can see that Re((zϕ2(t))ϕ2(t))=1+2s1+s12s1s>0whenevers12.

This shows that ϕ2(t) is convex in . Therefore, by Rogosinski’s lemma (Rogosinski, Citation1945), B1=s. Hence, using Theorem 2.6, we arrive at the desired result. □

Author contributions

All authors contributed equally to this work.

Ethical approval

Not applicable.

Acknowledgments

The authors are thankful to the Heads of their institutions for providing an excellent research environment.

Disclosure statement

No potential conflict of interest was reported by the authors.

Data availability statement

Not applicable.

Additional information

Funding

Not applicable.

References

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