Differential Subordinations of Arithmetic and Geometric Means of Some Functionals Related to a Sector

Abstract

Some general theorems on differential subordinations of some functionals connected with arithmetic and geometric means related to a sector are proved. These results unify a number of well known results concerning inclusion relation between the classes of analytic functions built with using arithmetic and geometric means.

1. Introduction

For 𝑟>0 let𝔻_𝑟={𝑧∈ℂ∶|𝑧|<𝑟}.Let𝔻=𝔻₁.

Let the functions𝑓and𝐹be analytic in the unit disc 𝔻. A function𝑓is called subordinate to𝐹,written 𝑓≺𝐹, if𝐹is univalent in 𝔻,𝑓(0)=𝐹(0) and 𝑓(𝔻)⊂𝐹(𝔻).

Let𝐷be a domain in ℂ² and 𝜓∶ℂ²⊃𝐷→ℂ be an analytic function, and let 𝑝 be a function analytic in 𝔻 with (𝑝(𝑧),𝑧𝑝^′(𝑧))∈𝐷,𝑧∈𝔻 andℎbe a function analytic and univalent in 𝔻. The function𝑝is said to satisfy the first-order differential subordination if 𝜓𝑝(𝑧),𝑧𝑝^′(𝑧)≺ℎ(𝑧),𝜓𝑝(0),0=ℎ(0),𝑧∈𝔻. (1.1)

The general theory of the differential subordinations has been studied intensively by many authors. A survey of this theory can by found in the monograph by Miller and Mocanu [1].

For 𝛽∈(0,2] let ℎ_𝛽(𝑧)=1+𝑧1−𝑧^𝛽,𝑧∈𝔻. (1.2) It is clear thatℎ_𝛽maps univalently 𝔻 onto the sector of the angle𝛽𝜋symmetrical with respect to the real axis with the vertex at the origin.

In this paper we are interested in the following problem referring to (1.1) to find the constant 𝑐_𝑘(𝑛,𝛾,𝛼,𝛽) so that to the following relation is true: 𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝^𝑘(𝑧)^𝛾≺ℎ_{𝑐𝑘(𝑛,𝛼,𝛾,𝛽)}(𝑧),𝑧∈𝔻⟹𝑝≺ℎ_𝛽, (1.3) with suitable assumptions on function𝑝and constants 𝑛,𝛼,𝛾,𝛽.For selected parameters 𝑛,𝛾,𝛼,𝛽 the theorems presented here reduce to the well-known theorems proved by various authors. Particularly, results of this type can be applied to examine inclusion relation between subclasses of analytic functions defined with using arithmetic or geometric means of some functionals, for example, the class of𝛼-convex functions or 𝛾-starlike functions.

The lemma below that slightly generalizes a lemma proved by Miller and Mocanu [2] will be required in our investigation.

Lemma 1.1 (see [2]). Let 𝑞∶𝔻→ℂ be a function analytic and univalent on 𝔻, injective on 𝜕𝔻 and 𝑞(0)=1. Let 𝑝(𝑧)=1+^∞_𝑘=𝑛𝑐_𝑘𝑧^𝑘,𝑧∈𝔻, (1.4) be analytic in 𝔻,𝑝≢1. Suppose that there exists a point 𝑧₀∈𝔻 such that 𝑝(𝑧₀)∈𝜕𝑞(𝔻) and 𝑝𝑧∈ℂ∶|𝑧|<𝑧₀⊂𝑞(𝔻). (1.5) If𝜉₀=𝑞⁻¹(𝑝(𝑧₀))and 𝑞^′(𝜉₀) exists, then there exists an𝑚≥𝑛for which 𝑧₀𝑝^′𝑧₀=𝑚𝜉₀𝑞^′𝜉₀. (1.6)

2. Main Results

In the first theorem which follows directly from Theorem 2.2 [3] we prove that𝑐_𝑘(𝑛,𝛼,𝛾,𝛽)≥𝛽.Let us start with the following definition.

Definition 2.1 (see [3]). Let𝛾∈[0,1]andΦbe a function analytic in domain 𝐷⊂ℂ. Byℋ(𝛾,Φ)will be denoted the class of functions𝑝analytic in 𝔻 with 𝑝(0)=1,𝑝≢1 and 𝑝(𝔻)⊂𝐷 such that the function 𝑄(𝑧)=𝑝(𝑧)1+𝑧𝑝′(𝑧)𝑝(𝑧)Φ(𝑝(𝑧))^𝛾,𝑄(0)=1,𝑧∈𝔻, (2.1) is well defined in 𝔻.

Theorem 2.2 (see [3]). Let 𝛾∈[0,1],ℎ a convex function such that 0∈ℎ(𝔻),ℎ(0)=1,Φ a function analytic in a domain 𝐷⊂ℂ such that ℎ(𝔻)⊂𝐷, and ReΦ(ℎ(𝑧))>0 for 𝑧∈𝔻. If 𝑝∈ℋ(𝛾,Φ)and 𝑝(𝑧)1+𝑧𝑝^′(𝑧)𝑝(𝑧)Φ𝑝(𝑧)^𝛾≺ℎ(𝑧),𝑧∈𝔻, (2.2) then 𝑝≺ℎ. (2.3)

Definition 2.3. Let𝑘,𝑛∈ℕ,𝛼≥0and𝛾∈[0,1].Byℋ_𝑘(𝑛,𝛼,𝛾)will be denoted the class of functions𝑝analytic in𝔻of the form (1.4) such that the function 𝑄(𝑧)=𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝^𝑘(𝑧)^𝛾,𝑄(0)=1,𝑧∈𝔻, (2.4) is well defined in𝔻.

Remark 2.4. (1) Setting Φ(𝑤)=Φ_𝑘,𝛼(𝑤)=𝛼𝑤^𝑘−1,𝑤∈ℂ⧵{0}, (2.5) we see that ℋ_𝑘1,𝛼,𝛾=ℋ𝛾,Φ_𝑘,𝛼. (2.6)
(2) For each𝑘,𝑛,𝛼,𝛾as in Definition 2.3 the classℋ_𝑘(𝑛,𝛼,𝛾)is nonempty. To see this take 𝑝(𝑧)=1+𝑐_𝑛𝑧^𝑛,𝑧∈𝔻, (2.7) for sufficiently small𝑐_𝑛∈ℂ.
(3) Clearly, for𝛾=0the classℋ_𝑘(𝑛,𝛼,0)contains all analytic functions𝑝of the form (1.4).
(4) Let𝑘=𝛾=1.Then 𝑄(𝑧)=𝑝(𝑧)+𝛼𝑧𝑝^′(𝑧),𝑧∈𝔻. (2.8) Therefore the classℋ₁(𝑛,𝛼,1)contains all analytic functions𝑝of the form (1.4).
(5) Let𝑝be analytic function in𝔻of the form (1.4). Suppose that𝑝(𝑧₀)=0for some𝑧₀∈𝔻.Then 𝑝(𝑧)=𝑧−𝑧₀^𝑚𝑝₁(𝑧),𝑧∈𝔻, (2.9) where𝑚≥1and𝑝₁is analytic function in𝔻with𝑝₁(𝑧₀)≠0for𝑧∈𝔻.Then we have 𝑧𝑝^′(𝑧)𝑝^𝑘(𝑧)𝑚=𝑧𝑧−𝑧₀^𝑚−1𝑝₁(𝑧)+𝑧−𝑧₀^𝑚𝑝^′₁(𝑧)𝑧−𝑧₀^𝑚𝑝₁(𝑧)^𝑘=𝑚𝑝₁(𝑧)+𝑧−𝑧₀𝑝^′₁(𝑧)𝑧−𝑧₀^{𝑚(𝑘−1)+1}𝑝^𝑘₁(𝑧). (2.10) Hence we see that for𝑘=1and𝛾∈(0,1)or for𝑘≥2and𝛾∈(0,1]the function 𝑄(𝑧)=𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝^𝑘(𝑧)^𝛾,𝑧∈𝔻, (2.11) has a pole at𝑧₀.Therefore for such𝑘and𝛾we see that every𝑝∈ℋ_𝑘(𝑛,𝛼,𝛾)is nonvanishing in𝐷.

Theorem 2.5. Let 𝑘∈ℕ,𝛼≥0,𝛾∈[0,1]and 𝛽∈[0,1]be such that (𝑘−1)𝛽≤1.If 𝑝∈ℋ_𝑘(1,𝛼,𝛾)and 𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝^𝑘(𝑧)^𝛾≺ℎ_𝛽(𝑧),𝑧∈𝔻, (2.12) then 𝑝≺ℎ_𝛽. (2.13)

Proof. The case𝛼=0is evident so we assume that𝛼>0.For𝑘∈𝑁and𝛼>0letΦ=Φ_𝑘,𝛼be defined by (2.5). For𝛽∈[0,1]the functionℎ_𝛽is convex with0∈𝜕ℎ(𝔻).Since(𝑘−1)𝛽≤1,we have ReΦℎ_𝛽(𝑧)=ReΦ1+𝑧1−𝑧^𝛽=𝛼Re1−𝑧1+𝑧^{(𝑘−1)𝛽}>0,𝑧∈𝔻. (2.14) Applying Theorem 2.2 withℎ_𝛽instead ofℎwe get the assertion.

Now we prove two theorems were we improve the result of Theorem 2.5. The problem (1.3) will be divided into two cases:𝑘=1and𝑘>1.

First we consider the case𝑘=1.The theorem below was proved in [4]. To be self-contained we include its proof.

Theorem 2.6. Fix 𝑛∈ℕ,𝛼≥0and 𝛾∈[0,1].Let 𝛽∈(0,𝛽₁(𝑛,𝛼,𝛾)],where 𝛽=𝛽₁(𝑛,𝛼,𝛾)is the solution of the equation 𝛽+𝑐₁𝑛,𝛼,𝛾,𝛽=4−𝛾, (2.15) with 𝑐₁𝑛,𝛼,𝛾,𝛽=𝛽+2𝛾𝜋arctan𝑛𝛼𝛽. (2.16) If𝑝∈ℋ₁(𝑛,𝛼,𝛾)and 𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝(𝑧)^𝛾≺ℎ_{𝑐1(𝑛,𝛼,𝛾,𝛽)}(𝑧),𝑧∈𝔻, (2.17) then 𝑝≺ℎ_𝛽. (2.18)

Proof. (1) Assume that 𝛼>0 and 𝛾∈(0,1] since the cases 𝛾=0 or 𝛼=0 are evident.
Suppose, on the contrary, that𝑝is not subordinate to ℎ_𝛽. Then, by the minimum principle for harmonic mappings there exists 𝑟₀∈(0,1) such that 𝑝𝔻_𝑟0⊂ℎ_𝛽(𝔻), (2.19) and one of the following cases hold: maxArg𝑝(𝑧)∶𝑧∈𝔻_𝑟0=maxArg𝑝(𝑧)∶|𝑧|=𝑟₀𝜋=𝛽2, (2.20) or minArg𝑝(𝑧)∶𝑧∈𝔻_𝑟0=minArg𝑝(𝑧)∶|𝑧|=𝑟₀𝜋=−𝛽2, (2.21) or 𝑝𝑧₀=0, (2.22) for some𝑧₀∈𝜕𝔻_𝑟0.
(2) Assume that (2.20) holds. Then there exists𝑧₀∈𝜕𝔻_𝑟0such that Arg𝑝𝑧₀𝜋=𝛽2. (2.23) Let𝜉₀=ℎ_𝛽⁻¹(𝑝(𝑧₀)).Thus 𝑝𝑧₀=ℎ_𝛽𝜉₀=1+𝜉₀1−𝜉₀^𝛽≠0. (2.24) Therefore𝜉₀≠±1and 1+𝜉₀1−𝜉₀=𝑥𝑖 (2.25) for𝑥>0,that is, 𝜉₀=𝑥𝑖−1𝑥𝑖+1. (2.26) Since𝜉₀≠±1,soℎ^′_𝛽(𝜉₀)exists. Hence and by Lemma 1.1 there exists an𝑚≥𝑛for which 𝑧₀𝑝^′𝑧₀=𝑚𝜉₀ℎ^′_𝛽𝜉₀. (2.27)
(3) Consequently, 𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝𝑧₀^𝛾=ℎ_𝛽𝜉₀𝜉1+𝑚𝛼⁰ℎ^′_𝛽𝜉₀ℎ_𝛽𝜉₀^𝛽=(𝑥𝑖)^𝛽1+𝑚𝛼𝛽1+𝑥²𝑖2𝑥^𝛾. (2.28) In view of the fact that𝑥>0let us take arg1+𝑚𝛼𝛽1+𝑥²𝑖∈𝜋2𝑥0,2. (2.29) Hence and from (2.28) we have arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝𝑧₀^𝛾=arg(𝑥𝑖)^𝛽1+𝑚𝛼𝛽1+𝑥²𝑖2𝑥^𝛾𝜋=𝛽2+𝛾arg1+𝑚𝛼𝛽1+𝑥²𝑖𝜋2𝑥=𝛽2+𝛾arctan𝑚𝛼𝛽1+𝑥².2𝑥 (2.30) By the above and by the fact that𝑚≥𝑛we have arg𝑝𝑧₀𝑧1+𝛼⁰𝑝′(𝑧₀)𝑝(𝑧₀)^𝛾𝜋≥𝛽2+𝛾arctan𝑛𝛼𝛽1+𝑥²𝜋2𝑥≥𝛽2+𝛾arctan𝑛𝛼𝛽=𝑐₁𝜋𝑛,𝛼,𝛾,𝛽2. (2.31) On the other hand, (2.30) yields arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝𝑧₀^𝛾𝜋=𝛽2+𝛾arctan𝑚𝛼𝛽1+𝑥²𝑖≤𝜋2𝑥𝛽+𝛾2. (2.32) Finally, the above and (2.31) lead to 𝑐₁𝜋𝑛,𝛼,𝛾,𝛽2≤arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝𝑧₀^𝛾≤𝜋𝛽+𝛾2≤2𝜋−𝑐₁𝜋𝑛,𝛼,𝛾,𝛽2, (2.33) for all𝛽∈(0,𝛽₁(𝑛,𝛼,𝛾)].
Thus we arrive at a contradiction with (2.17) so𝑝≺ℎ_𝛽.
(4) When (2.21) holds, we see that𝑥<0in (2.25). Next we finish the proof by similar argumentations like in the above.
(5) Assume now that (2.22) holds. In view of Remark 2.4 this is possible only when𝑘=𝛾=1.(a)For 𝛽<1 the boundary 𝜕ℎ_𝛽(𝔻) has the corner at 0 of the angle 𝛽𝜋<𝜋. Since 𝑝(𝜕𝔻_𝑟0) is an analytic curve, in view of (2.19) the case 𝑝(𝑧₀)=0 does not hold for 𝛽<1.(b)Let now 𝛽≥1.
Assume that𝑝^′(𝑧₀)≠0.Since𝑧₀𝑝^′(𝑧₀)is an outer normal to the curve𝑝(𝜕𝔻_𝑟0)at𝑝(𝑧₀), by (2.19) we see that 𝜋3−𝛽2=32𝜋𝜋−𝛽2≤arg𝑧₀𝑝^′𝑧₀𝜋≤𝛽2+𝜋2=𝜋𝛽+12. (2.34) Hence taking into account that 3−𝛽≤𝑐₁𝑛,𝛼,1,𝛽,𝛽+1≤4−𝑐₁,𝑝𝑧𝑛,𝛼,1,𝛽₀+𝛼𝑧₀𝑝^′𝑧₀=𝛼𝑧₀𝑝^′𝑧₀, (2.35) we deduce that 𝑐₁𝜋𝑛,𝛼,1,𝛽2≤arg𝑝𝑧₀+𝛼𝑧₀𝑝^′𝑧₀≤𝜋𝛽+𝛾2≤2𝜋−𝑐₁𝜋𝑛,𝛼,1,𝛽2, (2.36) for all𝛽∈(0,𝛽₁(𝑛,𝛼,𝛾)].In this way we arrive at a contradiction with (2.17) so𝑝≺ℎ_𝛽.
If𝑝^′(𝑧₀)=0,then 𝑝𝑧₀+𝛼𝑧₀𝑝^′𝑧₀=0∉ℎ_𝛽(𝔻), (2.37) and once again we contradict (2.17).

Special Cases
(1) The case𝑛=1,𝛼=1was proved in [5].
(2) The case𝛾=1was proved in [6].
Corollary 2.7 (see [6]). Let 𝑛∈ℕ,𝛼≥0.Let 𝛽∈(0,𝛽₁(𝑛,𝛼,1)],where 𝛽=𝛽₁(𝑛,𝛼,1)is the solution of the equation 𝛽+𝑐₁𝑛,𝛼,1,𝛽=3, (2.38) with 𝑐₁2𝑛,𝛼,1,𝛽=𝛽+𝜋arctan𝑛𝛼𝛽. (2.39) If𝑝is analytic function in𝔻of the form (1.4) and 𝑝(𝑧)+𝛼𝑧𝑝^′(𝑧)≺ℎ_{𝑐1(𝑛,𝛼,1,𝛽)}(𝑧),𝑧∈𝔻, (2.40) then 𝑝≺ℎ_𝛽. (2.41)(3)The case𝛾=1,𝛼=1was remarked [7].(4)The case𝛾=1,𝛼=1,𝑛=1was proved in detail in [8].

Now we consider the problem (1.3) for𝑘≥2.

Theorem 2.8. Let 𝑘≥2,𝑛∈ℕ,𝛼≥0,𝛾∈[0,1]and let 𝛽∈(0,1/(𝑘−1)].If 𝑝∈ℋ_𝑘(𝑛,𝛼,𝛾)and 𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝^𝑘(𝑧)^𝛾≺ℎ_{𝑐𝑘(𝑛,𝛼,𝛾,𝛽)}(𝑧),𝑧∈𝔻, (2.42) then 𝑝≺ℎ_𝛽, (2.43) where 𝑐_𝑘𝑛,𝛼,𝛾,𝛽=𝛽+2𝛾𝜋)×arctan𝑛𝛼𝛽cos(𝑘−1𝛽𝜋/21+(𝑘−1)𝛽^{(1+(𝑘−1)𝛽)/2}1−(𝑘−1)𝛽^{(1−(𝑘−1)𝛽)/2}().+𝑛𝛼𝛽sin𝑘−1𝛽𝜋/2 (2.44)

Proof. (1) We repeat argumentation from Parts 1 and 2 of the proof of Theorem 2.6.
(2) We have 𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝^𝑘𝑧₀^𝛾=ℎ_𝛽𝜉₀𝜉1+𝑚𝛼⁰ℎ^′_𝛽𝜉₀ℎ^𝑘_𝛽𝜉₀^𝛽=(𝑥𝑖)^𝛽1+𝑚𝛼𝛽1+𝑥²𝑖2𝑥(𝑥𝑖)^{(𝑘−1)𝛽𝛾}=(𝑥𝑖)^𝛽1+𝑚𝛼𝛽1+𝑥²2𝑥^{1+(𝑘−1)𝛽}𝑖^{1−(𝑘−1)𝛽𝛾}. (2.45) Since𝑥>0and1−(𝑘−1)𝛽≥0, we can take arg1+𝑚𝛼𝛽1+𝑥²2𝑥^{1+(𝑘−1)𝛽}𝑖^{1−(𝑘−1)𝛽}∈𝜋0,2. (2.46) Hence arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝^𝑘𝑧₀^𝛾=arg(𝑥𝑖)^𝛽1+𝑚𝛼𝛽1+𝑥²2𝑥^{1+(𝑘−1)𝛽}𝑖^{1−(𝑘−1)𝛽𝛾}𝜋=𝛽2+𝛾arg1+𝑚𝛼𝛽1+𝑥²2𝑥^{1+(𝑘−1)𝛽}()sin𝑘−1𝛽𝜋2+𝑖𝑚𝛼𝛽1+𝑥²2𝑥^{1+(𝑘−1)𝛽}()cos𝑘−1𝛽𝜋2𝜋=𝛽2+𝛾arctan𝑚𝛼𝛽1+𝑥²/2𝑥^{1+(𝑘−1)𝛽}cos(𝑘−1)𝛽𝜋/21+𝑚𝛼𝛽1+𝑥²/2𝑥^{1+(𝑘−1)𝛽}().sin𝑘−1𝛽𝜋/2 (2.47) Thus, from (2.46) and by the fact that𝑚≥𝑛we obtain arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝^𝑘𝑧₀^𝛾𝜋≥𝛽2(𝑥)()+𝛾arctan𝑛𝛼𝛽𝑎cos𝑘−1𝛽𝜋/2(𝑥)(),1+𝑛𝛼𝛽𝑎sin𝑘−1𝛽𝜋/2 (2.48) where 𝑎(𝑥)=1+𝑥²2𝑥^{1+(𝑘−1)𝛽},𝑥≠0. (2.49) We have 𝑎^′(𝑥)=1−(𝑘−1)𝛽𝑥²−1+(𝑘−1)𝛽2𝑥^{2+(𝑘−1)𝛽},𝑥≠0. (2.50)
(3) Assume now that(𝑘−1)𝛽<1.Observe that the function𝑎attains its minimum at the point 𝑥₀=1+(𝑘−1)𝛽1−(𝑘−1)𝛽. (2.51) Moreover 𝑎𝑥₀=11+(𝑘−1)𝛽^{(1+(𝑘−1)𝛽)/2}1−(𝑘−1)𝛽^{(1−(𝑘−1)𝛽)/2}. (2.52) Hence, and from (2.48), we have arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝^𝑘𝑧₀^𝛾𝜋≥𝛽2𝑥+𝛾arctan𝑛𝛼𝛽𝑎₀cos(𝑘−1)𝛽𝜋/21+𝑛𝛼𝛽𝑎𝑥₀)𝜋sin(𝑘−1𝛽𝜋/2=𝛽2()+𝛾arctan𝑛𝛼𝛽cos𝑘−1𝛽𝜋/21+(𝑘−1)𝛽^{(1+(𝑘−1)𝛽)/2}1−(𝑘−1)𝛽^{(1−(𝑘−1)𝛽)/2}+𝑛𝛼𝛽sin(𝑘−1)𝛽𝜋/2=𝑐_𝑘𝜋𝑛,𝛼,𝛾,𝛽2. (2.53) On the other hand, using the fact that0≤(𝑘−1)𝛽𝜋/2<𝜋/2,from (2.47) we obtain arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝^𝑘𝑧₀^𝛾𝜋=𝛽2+𝛾arctan𝑚𝛼𝛽1+𝑥²/2𝑥^{1+(𝑘−1)𝛽}cos(𝑘−1)𝛽𝜋/21+𝑚𝛼𝛽1+𝑥²/2𝑥^{1+(𝑘−1)𝛽}()≤𝜋sin𝑘−1𝛽𝜋/2𝛽+𝛾2. (2.54) Finally, the above and (2.53) yield 𝑐_𝑘𝜋𝑛,𝛼,𝛾,𝛽2≤arg𝑝𝑧₀𝑧1+𝛼⁰𝑝^′𝑧₀𝑝^𝑘𝑧₀^𝛾≤𝜋𝛽+𝛾2≤2𝜋−𝑐_𝑘𝜋𝑛,𝛼,𝛾,𝛽2, (2.55) for all𝛽∈(0,1/(𝑘−1)].
Thus we arrive at a contradiction with (2.17) so𝑝≺ℎ_𝛽.
(4) For(𝑘−1)𝛽=1we have𝑐_𝑘(𝑛,𝛼,𝛾,𝛽)=𝛽.
This ends the proof of the theorem for the case𝑥>0.
(5) When (2.21) holds, we see that𝑥<0in (2.25). Next we finish the proof by similar argumentations like in the above.
(6) Since𝛽≤1/(𝑘−1)<1,arguing as in Part 5(a) of the proof of Theorem 2.6 we see that the case (2.22) does not hold.

Special Cases
(1)𝛽=1/(𝑘−1).
Then𝑐_𝑘(𝑛,𝛼,𝛾,𝛽)=𝛽.
Corollary 2.9. Let 𝑘≥2,𝑛∈ℕ,𝛼≥0and 𝛾∈[0,1].If 𝑝∈ℋ_𝑘(𝑛,𝛼,𝛾)and 𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝^𝑘(𝑧)^𝛾≺ℎ_1/(𝑘−1)(𝑧),𝑧∈𝔻, (2.56) then 𝑝≺ℎ_1/(𝑘−1). (2.57)
(2)𝑘=2,𝛽=1.
Corollary 2.10. Let 𝑛∈ℕ,𝛼≥0and 𝛾∈[0,1].If 𝑝∈ℋ₂(𝑛,𝛼,𝛾)and Re𝑝(𝑧)1+𝛼𝑧𝑝^′(𝑧)𝑝²(𝑧)^𝛾>0,𝑧∈𝔻, (2.58) then Re𝑝(𝑧)>0,𝑧∈𝔻. (2.59)
(3) The case𝑛=1,𝛼=1was proved in [9].
(4)𝛾=1.
Corollary 2.11. Let 𝑘≥2,𝑛∈ℕ,𝛼≥0and 𝛽∈(0,1/(𝑘−1)].If 𝑝is a function analytic in 𝔻of the form (1.4) nonvanishing in 𝔻and 𝑝(𝑧)+𝛼𝑧𝑝^′(𝑧)𝑝^𝑘−1(𝑧)≺ℎ_{𝑐𝑘(𝑛,𝛼,1,𝛽)}(𝑧),𝑧∈𝔻, (2.60) then 𝑝≺ℎ_𝛽, (2.61) where𝑐_𝑘(𝑛,𝛼,1,𝛽)is given by (2.44).
(5)𝛾=1,𝑘=2.

Corollary 2.12. Let 𝑛∈ℕ,𝛼≥0, and 𝛽∈(0,1].If 𝑝is a function analytic in 𝔻of the form (1.4) nonvanishing in 𝔻and 𝑝(𝑧)+𝛼𝑧𝑝^′(𝑧)𝑝(𝑧)≺ℎ_{𝑐2(𝑛,𝛼,1,𝛽)}(𝑧),𝑧∈𝔻, (2.62) where 𝑐₂2𝑛,𝛼,1,𝛽=𝛽+𝜋arctan𝑛𝛼𝛽cos𝛽𝜋/21+𝛽^(1+𝛽)/2(1−𝛽)^(1−𝛽)/2+𝑛𝛼𝛽sin𝛽𝜋/2, (2.63) then 𝑝≺ℎ_𝛽. (2.64)

(6) The case𝛾=1,𝑘=2,𝑛=1,𝛼=1was proved in [10]. The same result was reproved in [11] and once again in [12].

(7)𝛾=1,𝑘=2,𝛽=1.

Corollary 2.13. Let 𝑛∈ℕ,𝛼≥0,and 𝛽∈(0,1].If 𝑝is an analytic function in 𝔻of the form (1.4) nonvanishing in 𝔻and Re𝑝(𝑧)+𝛼𝑧𝑝^′(𝑧)𝑝(𝑧)>0,𝑧∈𝔻, (2.65) then Re𝑝(𝑧)>0,𝑧∈𝔻. (2.66)

3. Applications

All this type results can be applied in the theory of analytic functions. Some results concerning the inclusion relations between subclasses of analytic functions can be formulated.

Let𝒜(𝑛),𝑛∈𝑁,denote the class of functions of the form 𝑓(𝑧)=𝑧+^∞_𝑘=𝑛+1𝑎_𝑘𝑧^𝑘, (3.1) which is analytic in𝔻.For short, let𝒜=𝒜(1).Also let𝒮denote the class of all functions in𝒜which are univalent in𝔻.

To use theorems and corollaries listed in the previous section we put instead of the function𝑝some functionals over the class𝒜(𝑛),such as𝑝(𝑧)=𝑓(𝑧)/𝑧,𝑝(𝑧)=𝑧𝑓^′(𝑧)/𝑓(𝑧)or the others. In this way the inclusion relations between selected subclasses of analytic functions can be obtained.

3.1. Arithmetic Means

(I)   𝛾=1,𝑘=1.
(i)𝑝(𝑧)=𝑓(𝑧)/𝑧,𝑧∈𝔻,𝑓∈𝒜(𝑛),𝑛∈𝑁.
For𝑛∈ℕ,𝛼≥0,𝛽∈(0,1]letℛ_𝑛(𝛼,𝛽)denote class of functions𝑓∈𝒜(𝑛)such that ()𝑓(𝑧)1−𝛼𝑧+𝛼𝑓^′(𝑧)≺ℎ_𝛽(𝑧),𝑧∈𝔻, (3.2) or, equivalently, )𝑓(𝑧)arg(1−𝛼𝑧+𝛼𝑓^′(𝑧𝜋)<𝛽2,𝑧∈𝔻. (3.3)
Using Corollary 2.7 we have the following.
Corollary 3.1. Let 𝑛∈ℕ,𝛼≥0, and 𝛽∈(0,𝛽₁(𝑛,𝛼,1)].If 𝑓∈𝒜(𝑛)and )𝑓(𝑧)arg(1−𝛼𝑧+𝛼𝑓^′(𝑧)<𝑐₁𝜋𝑛,𝛼,1,𝛽2, (3.4) then 𝑓(𝑧)arg𝑧𝜋<𝛽2,𝑧∈𝔻. (3.5)
The above result we can write in the following form.Corollary 3.2. ℛ_𝑛𝛼,𝑐₁𝑛,𝛼,1,𝛽⊂ℛ_𝑛0,𝛽. (3.6)
(ii)𝑝(𝑧)=𝑓^′(𝑧),𝑧∈𝔻,𝑓∈𝒜(𝑛),𝑛∈ℕ
For𝑛∈ℕ,𝛼≥0,𝛽∈(0,1]let𝒯_𝑛(𝛼,𝛽)denote class of functions𝑓∈𝒜(𝑛)such that 𝑓^′(𝑧)+𝛼𝑧𝑓^′′(𝑧)≺ℎ_𝛽(𝑧),𝑧∈𝔻, (3.7) or, equivalently, arg𝑓^′(𝑧)+𝛼𝑧𝑓^′′(𝑧𝜋)<𝛽2,𝑧∈𝔻. (3.8)
Remark 3.3. (1) The class𝒯₁(1,1)was introduced in [13].
(2) The class𝒯₁(𝛼,1)coincides with the class𝐻(𝛼,1,−1)studied in [14].
Observe that𝑓∈𝒯_𝑛(𝛼,𝛽)if and only if𝑧𝑓′∈ℛ_𝑛(𝛼,𝛽).
Using Corollary 2.7 we have the following.
Corollary 3.4. Let 𝑛∈ℕ,𝛼≥0, and 𝛽∈(0,𝛽₁(𝑛,𝛼,1)].If 𝑓∈𝒜(𝑛)and arg𝑓^′(𝑧)+𝛼𝑧𝑓^′′(𝑧)<𝑐₁𝜋𝑛,𝛼,1,𝛽2,𝑧∈𝔻, (3.9) then arg𝑓^′(𝑧𝜋)<𝛽2,𝑧∈𝔻. (3.10)
Hence we have the following.
Corollary 3.5. 𝒯_𝑛𝛼,𝑐₁𝑛,𝛼,1,𝛽⊂𝒯_𝑛0,𝛽. (3.11)

(II)𝛾=1,𝑘=2.
(ii)𝑝(𝑧)=𝑧𝑓^′(𝑧)/𝑓(𝑧),𝑧∈𝔻,𝑓∈𝒜(𝑛),𝑛∈ℕ.
For𝑛∈ℕ,𝛼≥0,𝛽∈(0,1]letℳ_𝑛(𝛼,𝛽)denote class of functions𝑓∈𝒜(𝑛)such that𝑓(𝑧)𝑓′(𝑧)≠0for𝑧∈𝔻and (1−𝛼)𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)≺ℎ_𝛽(𝑧),𝑧∈𝔻, (3.12) or, equivalently, arg(1−𝛼)𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)𝜋<𝛽2,𝑧∈𝔻. (3.13)

Remark 3.6. (1) The class ℳ₁(𝛼,1), that is, the class of so-called 𝛼-convex functions was introduced by Mocanu [15].
(2) The class ℳ₁(0,1) is identical with the class 𝒮^∗ of starlike functions. The classℳ₁(1,1)is identical with the class𝒞of convex functions.
(3) The class ℳ₁(0,𝛽) denoted by𝒮^∗(𝛽)were defined by Brannan and Kirwan [16] and, independently, by Stankiewicz [17, 18]. Functions in this class are called strongly starlike of order 𝛽.
The class ℳ₁(1,𝛽) denoted by 𝒞^∗(𝛽) contains functions called strongly convex of order 𝛽.
Using Corollary 2.12 we have the following result proved by Marjono and Thomas [19].
Corollary 3.7. Let 𝑛∈ℕ,𝛼≥0, and 𝛽∈(0,1].If 𝑓∈𝒜(𝑛)and arg(1−𝛼)𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)<𝑐₂𝜋𝑛,𝛼,1,𝛽2,𝑧∈𝔻, (3.14) then arg𝑧𝑓^′(𝑧)𝑓(𝑧)𝜋<𝛽2,𝑧∈𝔻. (3.15)
For𝛼=1one has the result due to Nunokawa and Thomas [12]:Corollary 3.8. Let 𝑛∈ℕand 𝛽∈(0,1].If 𝑓∈𝒜(𝑛)and arg1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)<𝑐₂𝜋𝑛,1,1,𝛽2,𝑧∈𝔻, (3.16) then arg𝑧𝑓^′(𝑧)𝑓(𝑧)<𝛽,𝑧∈𝔻. (3.17)Corollary 3.9. ℳ_𝑛𝛼,𝑐₂𝑛,𝛼,1,𝛽⊂ℳ_𝑛,ℳ0,𝛽₁(𝛼,1)⊂𝒮^∗,𝒞⊂𝒮^∗. (3.18)

3.2. Geometric Mean

(I)   𝛼=1,𝑘=1.
(i) 𝑝(𝑧)=𝑓(𝑧)/𝑧,𝑧∈𝔻,𝑓∈𝒜(𝑛),𝑛∈ℕ.
For𝑛∈ℕ,𝛽∈(0,1], and𝛾∈[0,1]letℒ_𝑛(𝛾,𝛽)denote class of functions𝑓∈𝒜(𝑛)such that 𝑓(𝑧)𝑧^1−𝛾𝑓^′(𝑧)^𝛾≺ℎ_𝛽(𝑧),𝑧∈𝔻, (3.19) or equivalently 𝑓(𝑧)arg𝑧^1−𝛾𝑓^′(𝑧)^𝛾𝜋<𝛽2,𝑧∈𝔻. (3.20)
Remark 3.10. The classℒ₁(𝛾,1)was introduced in [20].
Applying Theorem 2.6 with𝛼=1we have the following.
Corollary 3.11. Let𝑛∈ℕ,𝛾∈[0,1], and𝛽∈(0,𝛽₁(𝑛,1,𝛾)].If𝑓∈𝒜(1) and 𝑓(𝑧)arg𝑧^1−𝛾𝑓^′(𝑧)^𝛾<𝑐₁𝜋𝑛,1,𝛾,𝛽2,𝑧∈𝔻, (3.21) then 𝑓(𝑧)arg𝑧𝜋<𝛽2,𝑧∈𝔻. (3.22)Corollary 3.12. ℒ_𝑛𝛾,𝑐₁𝑛,1,𝛾,𝛽⊂ℒ_𝑛0,𝛽. (3.23)

(II)   𝛼=1,𝑘=2.
(ii)𝑝(𝑧)=𝑧𝑓^′(𝑧)/𝑓(𝑧),𝑧∈𝔻,𝑓∈𝒜(𝑛),𝑛∈ℕ.
For𝑛∈ℕ,𝛽∈(0,1], and𝛾∈[0,1]let𝒮^∗_𝑛(𝛾,𝛽)denote class of functions𝑓∈𝒜(𝑛)such that 𝑧𝑓^′(𝑧)𝑓(𝑧)^1−𝛾1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)^𝛾≺ℎ_𝛽(𝑧),𝑧∈𝔻, (3.24) or, equivalently, arg𝑧𝑓^′(𝑧)𝑓(𝑧)^1−𝛾1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)^𝛾𝜋<𝛽2,𝑧∈𝔻. (3.25)
Remark 3.13. (1) The class𝒮^∗_𝑛(𝛾,1),that is, the class of so-called 𝛾-starlike functions was introduced by Lewandowski et al. [21].
(2) Clearly, 𝒮^∗₁(0,1)=𝒮^∗,𝒮^∗₁()𝒮1,1=𝒞,_∗10,𝛽=ℳ₁0,𝛽=𝒮^∗𝛽,𝒮^∗₁1,𝛽=ℳ₁1,𝛽=𝒞^∗𝛽. (3.26)
Using Theorem 2.8 we obtain results due to Darus and Thomas [22].
Theorem 3.14. Let 𝑛∈ℕ,𝛾∈[0,1], and 𝛽∈(0,1].If 𝑓∈𝒜(𝑛)and arg𝑧𝑓^′(𝑧)𝑓(𝑧)^1−𝛾1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)^𝛾<𝑐₂𝜋𝑛,1,𝛾,𝛽2,𝑧∈𝔻, (3.27) then arg𝑧𝑓^′(𝑧)𝑓(𝑧)𝜋<𝛽2,𝑧∈𝔻. (3.28)Corollary 3.15. 𝒮^∗_𝑛𝛾;𝑐₂𝑛,1,𝛾,𝛽⊂𝒮^∗_𝑛𝛽,𝒮^∗₁𝛾,1⊂𝒮^∗. (3.29)
As further applications of Theorems 2.6 and 2.8 we can use arbitrary well-defined functionals over the class𝒜(𝑛).We recall two examples:
(1) 𝑝(𝑧)=𝐾_𝑎^𝛿+1𝑓(𝑧)𝑧,𝑧∈𝔻,𝑓∈𝒜(𝑛),𝑛∈ℕ,𝑎>0,𝛿≥0, (3.30) where the integral operator𝐾^𝛿_𝑎over the class𝒜(𝑛)was defined by Komatu [23] as follows: 𝐾^𝛿_𝑎𝑓(𝑧)=𝑎^𝛿Γ(𝛿)¹₀𝑡^𝑎−21log𝑡^𝛿−1𝑓(𝑧𝑡)𝑑𝑡,𝑧∈𝔻, (3.31) whereΓis the Gamma function;
(2) 𝑝(𝑧)=𝐿_𝜆𝑓(𝑧)𝑧,𝑧∈𝔻,𝑓∈𝒜,𝜆≥−1, (3.32) where the operator𝐿_𝜆over the class𝒜called Ruscheweyh derivative [24] was defined as follows: 𝐿_𝜆𝑓(𝑧)=𝑧(1−𝑧)^𝜆+1∗𝑓(𝑧),𝑧∈𝔻. (3.33)