*Article* **On a Harmonic Univalent Subclass of Functions Involving a Generalized Linear Operator**

#### **Abdeljabbar Talal Yousef and Zabidin Salleh \***

Department of Mathematics, Faculty of Ocean Engineering Technology and Informatics, Universiti Malaysia Terengganu, Kuala Nerus 21030, Terengganu, Malaysia; abduljabaryousef@gmail.com

**\*** Correspondence: zabidin@umt.edu.my

Received: 5 March 2020; Accepted: 21 March 2020; Published: 24 March 2020

**Abstract:** In this paper, a subclass of complex-valued harmonic univalent functions defined by a generalized linear operator is introduced. Some interesting results such as coefficient bounds, compactness, and other properties of this class are obtained.

**Keywords:** harmonic univalent functions; generalized linear operator; differential operator; Salagean operator; coefficient bounds

#### **1. Introduction**

Let *H* represent the continuous harmonic functions which are harmonic in the open unit disk *U* = {*z* : *z* ∈ C, |*z*| < 1} and let *A* be a subclass of *H* which represents the functions which are analytic in *U*. A harmonic function in *U* could be expressed as *f* = *h* + *g*, where *h* and *g* are in *A*, *h* is the analytic part of *f*, *g* is the co-analytic part of *f* and *h*- (*z*) > *g*- (*z*) is a necessary and sufficient condition for *<sup>f</sup>* to be locally univalent and sense-preserving in *U* (see Clunie and Sheil-Small [1]). Now we write,

$$h(z) = z + \sum\_{n=2}^{\infty} a\_n z^n,\\ g(z) = \sum\_{n=2}^{\infty} b\_n z^n. \tag{1}$$

Let *SH* represents the functions of the form *f* = *h* + *g* which are harmonic and univalent in *<sup>U</sup>*, which normalized by the condition *<sup>f</sup>*(0) = *fz*(0) <sup>−</sup> 1 = 0. The subclass *SH*<sup>0</sup> of *SH* consists of all functions in *SH* which have the additional property *fz*(0) = 0. The class *SH* was investigated by Clunie and Sheil-Smallas [1]. Since then, many researchers have studied the class *SH* and even investigated some subclasses of it. Also, we observe that the class *SH* reduces to the class *S* of normalized analytic univalent functions in *U*, if the co-analytic part of *f* is equal to zero. For *f* ∈ *S*, the Salagean differential operator *Dn*(*<sup>n</sup>* <sup>∈</sup> N<sup>0</sup> = N <sup>∪</sup> {0}) was defined by Salagean [2]. For *<sup>f</sup>* = *<sup>h</sup>* + *<sup>g</sup>* given by (1), Jahangiri et al. [3] defined the modified Salagean operator of *f* as

$$D^m f(z) = D^m h(z) + (-1)^m \overline{D^m g(z)}\_{\prime}$$

where

$$D^m h(z) = z + \sum\_{n=2}^{\infty} n^m a\_n z^n,\\ D^m g(z) = \sum\_{n=2}^{\infty} n^m b\_n z^n.$$

Next, for functions *f* ∈ *A*, For *n* ∈ N0, β ≥ γ ≥ 0, Yalçın and Altınkaya [4] defined the differential operator of *I m* <sup>γ</sup>,<sup>β</sup> *<sup>f</sup>* : *SH*<sup>0</sup> <sup>→</sup> *SH*<sup>0</sup> . Now we define our differential operator:

$$I\_{\delta,\mu,\lambda,\gamma,\varepsilon}^{0}f(z) = h(z) + \overline{g(z)}$$

$$I\_{\delta,\mu,\lambda,\varepsilon,\tau}^{1}f(z) = z + \sum\_{n=2}^{\infty} \left(\frac{\mu + \lambda - (\delta - \varepsilon)(\lambda - \tau)D^{0}f(z) + (\delta - \varepsilon)(\lambda - \tau)D^{1}f(z)}{\mu + \lambda}\right) \tag{2}$$

$$= \frac{\mu + \lambda - (\delta - \varepsilon)(\lambda - \tau)\left(h(z) + \overline{g(z)}\right) + (\delta - \varepsilon)(\lambda - \tau)\left(zh(z) + z\overline{g'(z)}\right)}{\mu + \lambda}$$

$$I\_{\delta,\mu,\lambda,\varepsilon,\tau}^{m}f(z) = I\_{\delta,\mu,\lambda,\varepsilon,\tau}^{1}\left(I\_{\delta,\mu,\lambda,\varepsilon,\tau}^{m-1}f(z)\right). \tag{3}$$

If *f* is given by (1), then from (2) and (3), we get (see [5])

$$\begin{split} \int\_{\delta,\mu,\lambda,\zeta,\tau}^{\mathrm{m}} f(z) &= z + \sum\_{n=2}^{\infty} \left( \frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda} \right)^{\mathrm{m}} a\_{\mathrm{n}} z^{\mathrm{n}} \\ &+ (-1)^{\mathrm{m}} \sum\_{n=2}^{\infty} \left( \frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda} \right)^{\mathrm{m}} \overline{b\_{\mathrm{n}} z^{\mathrm{n}}}. \end{split} \tag{4}$$

The operator *I m* <sup>δ</sup>,μ,λ,ς,<sup>τ</sup> *f*(*z*) generalizes the following differential operators:

If *f* ∈ *A*, then when we take μ = 1, λ = 0, δ = 0, τ = 1, ς = 1 we obtain *I m* 0,τ, <sup>δ</sup>,<sup>ς</sup> *f*(*z*) was introduced and studied by Ramadan and Darus [6]. By taking different choices of μ, λ, δ, τ *and* ς we get *I m* <sup>1</sup>−λ,τ, 0,<sup>ς</sup> *<sup>f</sup>*(*z*) was introduced and studied by Darus and Ibrahim [7], *<sup>I</sup> m* <sup>μ</sup>, <sup>λ</sup>,0, 1,0 *f*(*z*) was introduced and studied by Swamy [8], *I m* <sup>1</sup>−λ,0, 1,0 *<sup>f</sup>*(*z*) was introduced and studied by Al-Oboudi [9] and *<sup>I</sup> m* 0, 0,, 1,0 *f*(*z*) was introduced and studied by Salagean [2].

If *f* ∈ *H*, then *I m* <sup>μ</sup>, <sup>λ</sup>,0, 1,0 *f*(*z*) becomes the modified Salagean operator introduced by Yasar and Yalçin [10].

A function *f* : *U* → *C* is subordinate to the function *g* : *U* → *C* denoted by *f*(*z*) ≺ *g*(*z*), if there exists an analytic function *w* : *U* → *U* with *w*(0) = 0 such that

$$f(z) = \lg(w(z)), (z \in \mathcal{U})\,\_{\ast}$$

Moreover, if the function *g* is univalent in *U*, then we have (see [11,12]):

$$f(z) \preccurlyeq \mathcal{g}(z) \text{ if and only if } f(0) = \mathcal{g}(0), \ f(\mathcal{U}) \subset \mathcal{g}(\mathcal{U}).$$

Denote by *SH*0(δ, μ, λ, ς, τ, *m*, *A*, *B*) the subclass of *SH*<sup>0</sup> consisting of functions of the form (1) that satisfy the condition

$$\frac{I\_{\delta,\mu,\lambda,\varsigma,\pi}^{m+1}f(z)}{I\_{\delta,\mu,\lambda,\varsigma,\pi}^{m}f(z)} < \frac{1+Az}{1+Bz}, -1 \le A < B \le 1\tag{5}$$

where *I m* <sup>δ</sup>,μ,λ,η,ς,<sup>τ</sup> *f*(*z*) is defined by (4). For relevant and recent references related to this work, we refer the reader to [13–20].

In this paper we use the same techniques that have been used earlier by Dziok [21] and Dziok et al. [22], to investigate coefficient bound, distortion bounds, and some other properties for the class *SH*0(δ, μ, λ, ς, τ, *m*, *A*, *B*).

#### **2. Coe**ffi**cient Bounds**

In this section we find the coefficient bound for the class *SH*0(δ, μ, λ, ς, τ, *m*, *A*, *B*).

**Theorem 1.** *Let the function f*(*z*) = *<sup>h</sup>* + *g be defined by (1). Then f* <sup>∈</sup> *SH*0(δ, <sup>μ</sup>, <sup>λ</sup>, <sup>ς</sup>, <sup>τ</sup>, *<sup>m</sup>*, *<sup>A</sup>*, *<sup>B</sup>*) if

$$\sum\_{n=2}^{\infty} \left( \mathbb{C}\_n |a\_n| + D\_n |b\_n| \right) \le B - A \tag{6}$$

*where*

$$\mathbf{C}\_{n} = \left(\frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda}\right)^{m} \left\{\frac{(\delta - \varepsilon)(\lambda - \tau)(n - 1)[B + 1] - (\mu + \lambda)(B - A)}{\mu + \lambda}\right\} \tag{7}$$

*and*

$$D\_n = \left(\frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda}\right)^m \left\{\frac{[A + B(2 + (\delta - \varepsilon)(\lambda - \tau)(n - 1))](\mu + \lambda)}{\mu + \lambda}\right\}.\tag{8}$$

**Proof.** Let *an* - 0 or *bn* -0 for *n* ≥ 2. Since *Cn*, *Dn* ≥ *n*(*B* − *A*) by (6), we obtain

$$\begin{aligned} \left| h'(z) \right| - \left| g'(z) \right| &\geq 1 - \sum\_{n=2}^{\infty} n |a\_n| |z|^{n-1} - \sum\_{n=2}^{\infty} n |b\_n| |z|^{n-1} \\ &\geq 1 - |z| \sum\_{n=2}^{\infty} \left( n |a\_n| + n |b\_n| \right) \\ &\geq 1 - \frac{|z|}{B - A} \sum\_{n=2}^{\infty} \left( \mathbb{C}\_n |a\_n| + D\_n |b\_n| \right) \\ &\geq 1 - |z| > 0. \end{aligned}$$

Therefore, *f* is univalent in *U*. To ensure the univalence condition, consider *z*1, *z*<sup>2</sup> ∈ *U* so that *z*<sup>1</sup> *z*2. Then

$$\left| \frac{z\_1^n - z\_2^n}{z\_1 - z\_2} \right| = \left| \sum\_{m=1}^n z\_1^{m-1} - z\_2^{n-m} \right| \le \sum\_{m=1}^n \left| z\_1^{m-1} \right| \left| z\_2^{n-m} \right| < n \text{ , } n \ge 2.1$$

So, we have

 

$$\left|\frac{f(z\_1) - f(z\_2)}{h(z\_1) - h(z\_2)}\right| \ge 1 - \left|\frac{\chi(z\_1) - \chi(z\_2)}{h(z\_1) - h(z\_2)}\right| = 1 - \left|\frac{\sum\_{n=2}^{\infty} h\_{\mathfrak{k}}\left(z\_1^n - z\_2^n\right)}{z\_1 - z\_2 + \sum\_{n=2}^{\infty} a\_{\mathfrak{k}}\left(z\_1^n - z\_2^n\right)}\right|$$

$$> 1 - \frac{\sum\_{n=2}^{\infty} n|b\_{\mathfrak{k}}|}{1 - \sum\_{n=2}^{\infty} n|a\_{\mathfrak{k}}|} \ge 1 - \frac{\sum\_{n=2}^{\infty} \frac{D\_{\mathfrak{k}}}{B - \mathfrak{k}} |b\_{\mathfrak{k}}|}{\sum\_{n=2}^{\infty} \frac{C\_{\mathfrak{k}}}{B - \mathfrak{k}} |a\_{\mathfrak{k}}|} \ge 0,$$

which proves univalences.

On the other hand, *<sup>f</sup>* <sup>∈</sup> *SH*0(δ, <sup>μ</sup>, <sup>λ</sup>, <sup>ς</sup>, <sup>τ</sup>, *<sup>m</sup>*, *<sup>A</sup>*, *<sup>B</sup>*) if and only if there exists a function *<sup>w</sup>*; with *w*(0) = 0, and *w*(*z*) <sup>&</sup>lt; <sup>1</sup>(*<sup>z</sup>* <sup>∈</sup> *<sup>U</sup>*) such that

$$\frac{I\_{\delta,\mu,\lambda,\varsigma,\pi}^{m+1}f(z)}{I\_{\delta,\mu,\lambda,\varsigma,\pi}^{m}f(z)} < \frac{1+Az}{1+Bz}$$

or

$$\frac{I\_{\delta,\mu,\lambda,\varsigma,\tau}^{m+1}f(z) - I\_{\delta,\mu,\lambda,\varsigma,\tau}^{m}f(z)}{BI\_{\delta,\mu,\lambda,\varsigma,\tau}^{m+1}f(z) - AI\_{\delta,\mu,\lambda,\varsigma,\tau}^{m}f(z)} < 1, \quad (z \in \mathsf{U}).\tag{9}$$

The above inequality (9) holds, since for |*z*| = *r* (0 < *r* < 1) we obtain

 *I m*+1 <sup>δ</sup>,μ,λ,ς,<sup>τ</sup> *f*(*z*) − *I m* <sup>δ</sup>,μ,λ,ς,<sup>τ</sup> *f*(*z*) − *BIm*+<sup>1</sup> <sup>δ</sup>,μ,λ,ς,<sup>τ</sup> *<sup>f</sup>*(*z*) <sup>−</sup> *AI<sup>m</sup>* <sup>δ</sup>,μ,λ,ς,<sup>τ</sup> *f*(*z*) = ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *<sup>m</sup>* (δ−ς)(λ−τ)(*n*−1) <sup>μ</sup>+<sup>λ</sup> *anzn* +(−1) *<sup>m</sup>* ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *<sup>m</sup>* <sup>2</sup>(μ+λ)+(δ−ς)(λ−τ)(*n*−1) <sup>μ</sup>+<sup>λ</sup> *bnzn* − (*B* − *A*)*z* + ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *m <sup>B</sup>*μ+λ+(δ−ς)(λ−τ)(*n*−1) <sup>μ</sup>+<sup>λ</sup> − *A anzn* −(−1) *<sup>m</sup>* ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *m B*, 2(μ+λ)+δ(−ς)(λ−τ)(1−*n*) <sup>μ</sup>+<sup>λ</sup> + *A bnzn* ≤ ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *<sup>m</sup>* (δ−ς)(λ−τ)(*n*−1) <sup>μ</sup>+<sup>λ</sup> <sup>|</sup>*an*|*rn*+ ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *<sup>m</sup>* <sup>2</sup>(μ+λ)+(δ−ς)(λ−τ)(1−*n*) <sup>μ</sup>+<sup>λ</sup> <sup>|</sup>*bn*|*rn* <sup>−</sup> (*<sup>B</sup>* <sup>−</sup> *<sup>A</sup>*)*<sup>r</sup>* + ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *m <sup>B</sup>*μ+λ+(δ−ς)(λ−τ)(*n*−1)+*<sup>A</sup>* <sup>μ</sup>+<sup>λ</sup> − *A* <sup>|</sup>*an*|*rn* + ∞ *n*=2 μ+λ+(δ−ς)(λ−τ)(*n*−1) μ+λ *m <sup>B</sup>*2(μ+λ)+(δ−ς)(λ−τ)(*n*−1) <sup>μ</sup>+<sup>λ</sup> + *A* <sup>|</sup>*bn*|*rn* ≤ *r* - ∞ *n*=2 (*Cn*|*an*<sup>|</sup> + *Dn*|*bn*|)*rn*−<sup>1</sup> <sup>−</sup> (*<sup>B</sup>* <sup>−</sup> *<sup>A</sup>*) < 0.

Therefore, *<sup>f</sup>* <sup>∈</sup> *SH*0(δ, <sup>μ</sup>, <sup>λ</sup>, <sup>ς</sup>, <sup>τ</sup>, *<sup>m</sup>*, *<sup>A</sup>*, *<sup>B</sup>*), and so the proof is completed.

Next we show that the condition (6) is also necessary for the functions *f* ∈ *H* to be in the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, <sup>μ</sup>, <sup>λ</sup>, <sup>ς</sup>, <sup>τ</sup>, *<sup>m</sup>*, *<sup>A</sup>*, *<sup>B</sup>*) <sup>=</sup> *<sup>T</sup><sup>m</sup>* <sup>∩</sup> *SH*0(δ, <sup>μ</sup>, <sup>λ</sup>, <sup>ς</sup>, <sup>τ</sup>, *<sup>m</sup>*, *<sup>A</sup>*, *<sup>B</sup>*) where *<sup>T</sup><sup>m</sup>* is the class of functions *<sup>f</sup>* = *<sup>h</sup>* + *<sup>g</sup>* <sup>∈</sup> *SH*<sup>0</sup> so that

$$f = h + \overline{\chi} = z - \sum\_{n=2}^{\infty} a\_n z^n + (-1)^m \sum\_{n=2}^{\infty} |b\_n| \overline{z^n} (z \in \mathcal{U}).\tag{10}$$

 


**Theorem 2.** *Let <sup>f</sup>* = *<sup>h</sup>* + *<sup>g</sup> be defined by (10). Then <sup>f</sup>* <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) *if and only if the condition (6) holds.*

**Proof.** For this proof, we let the fractions (δ−ς)(λ−τ)(*n*−1) <sup>μ</sup>+<sup>λ</sup> <sup>=</sup> *<sup>L</sup>* and <sup>2</sup>(μ+λ)+(δ−ς)(λ−τ)(*n*−1) <sup>μ</sup>+<sup>λ</sup> = *K*. The first part "if statement" follows from Theorem 1. Conversely, we suppose that *<sup>f</sup>* <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*), then by (9) we have

$$\left| \frac{\sum\_{n=2}^{\infty} \left[ (L)^m \frac{(\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda} |a\_n| z^n + (K)^m \frac{2(\mu + \lambda) + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda} \overline{|b\_n| z^n} \right]}{(B - A)z - \sum\_{n=2}^{\infty} \left[ (L)^m (BL - A) |a\_n| z^n + (K)^m (BK + A) |b\_n| z^n \right]} \right| < 1.$$

For |*z*| = *r* < 1, we obtain

$$\frac{\sum\_{n=2}^{\infty} \left[ (L)^m \frac{(\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda} |a\_n| + (K)^m \frac{2(\mu + \lambda) + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda} |\overline{b\_n}| \right] r^{n - 1}}{(B - A) - \sum\_{n=2}^{\infty} \left[ (L)^m (BL - A) |a\_n| + (K)^m (BK + A) |b\_n| \right] r^{n - 1}} < 1.$$

Thus, for *Cn* and *Dn* as defined by (7) and (8), we have

$$\sum\_{n=2}^{\infty} \left[ C\_n |a\_n| + D\_n |b\_n| \right] r^{n-1} < B - A \left( 0 \le r < 1 \right). \tag{11}$$

Let ρ*n* be the sequence of partial sums of the series

$$\sum\_{k=2}^{m} \left[ C\_k |a\_k| + D\_k |b\_k| \right].$$

Then ρ*n* is a non-decreasing sequence and by (11) it is bounded above by *B* − *A*. Thus, it is convergent and

$$\sum\_{n=2}^{\infty} \left[ C\_n |a\_n| + D\_n |b\_n| \right] = \lim\_{n \to +\infty} \rho\_n \le B - A.$$

This gives us the condition (6). -

#### **3. Compactness and Convex**

In this section we obtain the compactness and the convex relation for the class *SH*0(δ, μ, λ, ς, τ, *m*, *A*, *B*).

**Theorem 3.** *The class SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) *is convex and compact subset of SH*.

**Proof.** Let *ft* <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*), where

$$f\_t(z) = z - \sum\_{n=2}^{\infty} |a\_{t,n}| z^n + (-1)^m \sum\_{n=2}^{\infty} |b\_{t,n}| \overline{z^n} (z \in \mathbb{U}, \ t \in \mathbb{N}).\tag{12}$$

Then for 0 <sup>≤</sup> <sup>ψ</sup> <sup>≤</sup> 1, let *<sup>f</sup>*1, *<sup>f</sup>*<sup>2</sup> <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) be defined by (12). Then

$$\begin{aligned} \xi(z) &= \psi f\_1(z) + (1 - \psi) f\_2(z) \\ &= z - \sum\_{n=2}^{\infty} \left( \psi \big| a\_{1,n} \big| + (1 - \psi) \big| a\_{2,n} \big| \right) z^n + (-1)^m \sum\_{n=2}^{\infty} \left( \psi \big| b\_{1,n} \big| + (1 - \psi) \big| b\_{2,n} \big| \right) \overline{z^n} \end{aligned}$$

and

$$\begin{split} &\sum\_{n=2}^{\infty} \left\{ \mathbb{C}\_{n} \Big( \psi \big| a\_{1,n} \big| + (1-\psi) \big| a\_{2,n} \big| \right) + D\_{n} \Big( \psi \big| b\_{1,n} \big| + (1-\psi) \big| b\_{2,n} \big| \Big) \right\} \\ &= \psi \sum\_{n=2}^{\infty} \left\{ \mathbb{C}\_{n} \Big| a\_{1,n} \big| + D\_{n} \Big| b\_{1,n} \big| \right\} + (1-\psi) \sum\_{n=2}^{\infty} \left\{ \mathbb{C}\_{n} \Big| a\_{2,n} \big| + D\_{n} \Big| b\_{2,n} \Big| \right\} \\ &\leq \psi (B-A) + (1-\psi)(B-A) = B-A. \end{split}$$

Thus, the function <sup>ξ</sup> = <sup>ψ</sup>*f*1(*z*) + (<sup>1</sup> <sup>−</sup> <sup>ψ</sup>)*f*2(*z*) is in the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*). This implies that *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) is convex.

For *ft* <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, <sup>μ</sup>, <sup>λ</sup>, <sup>ς</sup>, <sup>τ</sup>, *<sup>m</sup>*, *<sup>A</sup>*, *<sup>B</sup>*), *<sup>t</sup>* <sup>∈</sup> N and <sup>|</sup>*z*<sup>|</sup> <sup>≤</sup> *<sup>r</sup>* (<sup>0</sup> <sup>&</sup>lt; *<sup>r</sup>* <sup>&</sup>lt; <sup>1</sup>), then we have

$$\left|f\_{t}(z)\right| \le r + \sum\_{n=2}^{\infty} \left\{ \left|a\_{t,n}\right| + \left|b\_{t,n}\right| \right\} r^{n}$$

$$\le r + \sum\_{n=2}^{\infty} \left\{ \mathbb{C}\_{n} \left|a\_{t,n}\right| + D\_{n} \left|b\_{t,n}\right| \right\} r^{n}$$

$$\le r + (B - A)r^{2}.$$

Therefore, *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) is uniformly bounded. Let

$$f\_t(z) = z - \sum\_{n=2}^{\infty} \left| a\_{t,n} \right| z^n + (-1)^m \sum\_{n=2}^{\infty} \left| b\_{t,n} \right| \overline{z^n} (z \in \mathcal{U}, \ t \in \mathbb{N}).$$

also, let *f* = *h* + *g* where *h* and *g* are given by (1). Then by Theorem 2 we get

$$\sum\_{n=2}^{\infty} \left\{ \mathbb{C}\_n |a\_n| + D\_n \left| b\_{t,n} \right| \right\} \le B - A. \tag{13}$$

If we assume *ft* <sup>→</sup> *<sup>f</sup>*, then we get that *at*,*n* <sup>→</sup> <sup>|</sup>*an*<sup>|</sup> and *bt*,*n* <sup>→</sup> <sup>|</sup>*bn*<sup>|</sup> as *<sup>n</sup>* <sup>→</sup> <sup>+</sup><sup>∞</sup> (*<sup>t</sup>* <sup>∈</sup> <sup>N</sup>). Let ρ*n* be the sequence of partial sums of the series ∞ *n*=2 *Cn at*,*n* <sup>+</sup> *Dn bt*,*n* . Then ρ*n* is a non-decreasing sequence and by (13) it is bounded above by *B* − *A*. Thus, it is convergent and

$$\sum\_{n=2}^{\infty} \left\{ C\_n \left| a\_{t,n} \right| + D\_n \left| b\_{t,n} \right| \right\} = \lim\_{n \to \infty} \rho\_n \le B - A.$$

Therefore, *<sup>f</sup>* <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, <sup>μ</sup>, <sup>λ</sup>, <sup>ς</sup>, <sup>τ</sup>, *<sup>m</sup>*, *<sup>A</sup>*, *<sup>B</sup>*) and therefore the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) is closed. As a result, the class is closed, and the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) is also compact subset of *SH*, which completes the proof. -

**Lemma 1** [23]. *Let f* = *h* + *g be so that h and g are given by (1). Furthermore, let*

$$\sum\_{n=2}^{\infty} \left\{ \frac{n-\alpha}{1-\alpha} |a\_{\mathcal{U}}| + \frac{n+\alpha}{1-\alpha} |b\_{\mathcal{U}}| \right\} \le 1 (z \in \mathcal{U})$$

*where* 0 ≤ α < 1. *Then f is harmonic, orientation preserving, univalent in U and f is starlike of order* α.

**Theorem 4.** *Let* 0 ≤ α < 1, *Cn and Dn be defined by (7) and (8). Then*

$$r\_a^\* \left( S H\_T^0 (\delta\_\prime \mu, \lambda, \zeta, \tau, n, A, B) \right) = \inf\_{n \ge 2} \left[ \frac{1 - \alpha}{B - A} \min \left\{ \frac{C\_n}{n + \alpha'}, \frac{D\_n}{n + \alpha} \right\} \right]^{\frac{1}{n - 1}},\tag{14}$$

*where r*∗ <sup>α</sup> *is the radius of starlikeness of order* α. **Proof.** Let *<sup>f</sup>* <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) be of the form (10). Then, for |*z*| = *r* < 1, we get

$$\begin{split} & \left| \frac{I\_{0,\eta}f(z) - (1+\alpha)f(z)}{I\_{0,\eta}f(z) + (1+\alpha)f(z)} \right| \\ &= \left| \frac{-\alpha z - \sum\_{n=2}^{\infty} (n-1-\alpha)|a\_{n}|z^{n} - (-1)^{m} \sum\_{n=2}^{\infty} (n+1+\alpha)|b\_{n}|z^{n}}{(2-\alpha)z - \sum\_{n=2}^{\infty} (n-1-\alpha)|a\_{n}|z^{n} - (-1)^{m} \sum\_{n=2}^{\infty} (n-1+\alpha)|b\_{n}|z^{n}} \right| \\ &\leq \frac{\alpha - \sum\_{n=2}^{\infty} \left\{ (n-1-\alpha)|a\_{n}| - (-1)^{m} \sum\_{n=2}^{\infty} (n+1+\alpha)|b\_{n}| \right\} r^{n-1}}{2-\alpha - \sum\_{n=2}^{\infty} \left\{ (n-1-\alpha)|a\_{n}| - (-1)^{m} \sum\_{n=2}^{\infty} (n-1+\alpha)|b\_{n}| \right\}} .\end{split}$$

By using Lemma 1, we observe that *f* is starlike of order α in *Ur* if and only if

$$
\left|\frac{I\_{0,\eta}f(z) - (1+\alpha)f(z)}{I\_{0,\eta}f(z) + (1+\alpha)f(z)}\right| < 1, z \in \mathcal{U}\_r
$$

$$
\sum\_{n=2}^{\infty} \left(\frac{n-\alpha}{1-\alpha}|a\_n| + \frac{n+\alpha}{1-\alpha}|b\_n|\right) r^{n-1} \le 1. \tag{15}
$$

Furthermore, by using Theorem 2, we get

$$\sum\_{n=2}^{\infty} \left\{ \frac{C\_n}{1-\alpha} |a\_n| + \frac{D\_n}{1-\alpha} |b\_n| \right\} r^{n-1} \le 1.$$

Condition (15) is true if

$$\frac{n-\alpha}{1-\alpha}r^{n-1} \le \frac{C\_n}{B-A}r^{n-1}\dots$$

This proves

or

$$\frac{n+\alpha}{1-\alpha}r^{n-1} \le \frac{D\_n}{B-A}r^{n-1}(n=2,3\dots).$$

So, the function *f* is starlike of order α in the disk *U*∗ *<sup>r</sup>*<sup>α</sup> where

$$r\_\alpha^\* = \inf\_{n \ge 2} \left[ \frac{1-\alpha}{B-A} \min \left\{ \frac{C\_n}{n+\alpha'}, \frac{D\_n}{n+\alpha} \right\} \right]^{\frac{1}{n-1}},$$

and the function

$$f\_n(z) = h\_n(z) + \overline{g\_n(z)} = z - \frac{B - A}{\mathbb{C}\_n} z^n + (-1)^m \frac{B - A}{D\_n} \overline{z^n}.$$

So, the radius *r*∗ <sup>α</sup> cannot be larger. Then we get (14). -

#### **4. Extreme Points**

In this section we find the extreme points for the class *SH*0(δ, μ, λ, ς, τ, *m*, *A*, *B*).

**Theorem 5.** *The extreme points of SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) *are the functions f of the form (1) where h* = *hk and g* = *gk are of the form*

$$h\_1(z) = z,$$

$$h\_{ll}(z) = z - \frac{B - A}{C\_n} z^n,\tag{16}$$

$$g\_n(z) = (-1)^n \frac{B - A}{D\_n} \overline{z^n}, \ (z \in \mathcal{U}, \ n \ge 2).$$

**Proof.** Suppose that *gn* = <sup>ψ</sup>*f*<sup>1</sup> + (<sup>1</sup> <sup>−</sup> <sup>ψ</sup>)*f*<sup>2</sup> where 0 <ψ< 1 and *<sup>f</sup>*1, *<sup>f</sup>*<sup>2</sup> <sup>∈</sup> *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) are written in the form

$$f\_t(z) = z - \sum\_{n=2}^{\infty} \left| a\_{t,n} \right| z^n + (-1)^{\text{tr}} \sum\_{n=2}^{\infty} \left| b\_{t,n} \right| \overline{z^n} (z \in \mathcal{U}, \ t \in \{1, 2\}) \dots$$

Then, by (16), we get

$$\left| b\_{1,n} \right| = \left| b\_{2,n} \right| = \frac{B - A}{D\_n} \, ^\prime$$

and *a*1,*<sup>t</sup>* = *a*2,*<sup>t</sup>* = 0 for *t* ∈ {2, 3 ...} and *b*1,*<sup>t</sup>* = *b*2,*<sup>t</sup>* = 0 for *t* ∈ {2, 3 ...} - {*n*}. It follows that *gn*(*z*) = *f*1(*z*) = *f*2(*z*) and *gn* are in the class of extreme points of the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*). We also can ensure that the functions *hn*(*z*) are the extreme points of the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*). Now, assume that a function *f* of the form (1) is in the class of the extreme points of the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) and *f* is not of the form (16). Then there exists *k* ∈ {2, 3 ...} such that

$$|0 < |a\_k| < \frac{B - A}{\left(\frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(k - 1)}{\mu + \lambda}\right)^m \left\{\frac{(\delta - \varepsilon)(\lambda - \tau)[(k - 1)(B + 1)] + (\mu + \lambda)(B - A)}{\mu + \lambda}\right\}}$$

or

$$|0| < |b\_k| < \frac{B - A}{\left(\frac{\mu + \lambda - (\delta - \varepsilon)(\lambda - \tau)(n + 1)}{\mu + \lambda}\right)^m \left\{\frac{[A + B(2 + (\delta - \varepsilon)(\lambda - \tau)(n - 1))](\mu + \lambda)}{\mu + \lambda}\right\}^{-1}}$$

If

$$|0| < |a\_k| < \frac{B - A}{\left(\frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda}\right)^m \left\{\frac{(\delta - \varepsilon)(\lambda - \tau)(n - 1)[B + 1] - (\mu + \lambda)(B - A)}{\mu + \lambda}\right\}} $$

then putting

$$\psi = \frac{|a\_k| \left[ \left( \frac{\mu + \lambda + (\delta - \varsigma)(\lambda - \tau)(n - 1)}{\mu + \lambda} \right)^m \left\{ \frac{(\delta - \varsigma)(\lambda - \tau)(n - 1)[B + 1] - (\mu + \lambda)(B - A)}{\mu + \lambda} \right\} \right]}{B - A}$$

and

$$\chi = \frac{f - \psi h\_k}{1 - \psi} \prime$$

we have 0 <ψ< 1, *hk* χ. Therefore, *f* is not in the class of the extreme points of the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, η, ς, τ, *m*, *A*, *B*). Similarly, if

$$|0 < |b\_k| < \frac{B - A}{\left(\frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda}\right)^m \left\{\frac{[A + B(2 + (\delta - \varepsilon)(\lambda - \tau)(n - 1))](\mu + \lambda)}{\mu + \lambda}\right\}}$$

then putting

$$\psi = \frac{|\mathbb{B}\_{k}| \left( \frac{\mu + \lambda + (\delta - \varepsilon)(\lambda - \tau)(n - 1)}{\mu + \lambda} \right)^{\mathrm{II}} \left\{ \frac{[A + B(2 + (\delta - \varepsilon)(\lambda - \tau)(n - 1))](\mu + \lambda)}{\mu + \lambda} \right\}}{B - A}$$

$$\chi = \frac{f - \psi \mathbb{g}\_{k}}{1 - \psi} \,\,\,\,$$

and

we have 0 <ψ< 1, *gk* χ. It follows that *f* is not in the family of extreme points of the class *SH*<sup>0</sup> *<sup>T</sup>*(δ, μ, λ, ς, τ, *m*, *A*, *B*) and so the proof is completed. -

**Author Contributions:** Conceptualization, A.T.Y. and Z.S.; methodology, A.T.Y.; software, A.T.Y.; validation, A.T.Y. and Z.S.; formal analysis, A.T.Y.; investigation, A.T.Y.; resources, A.T.Y.; data curation, A.T.Y.; writing—original draft preparation, A.T.Y.; writing—review and editing, Z.S.; visualization, Z.S.; supervision, Z.S.; project administration, Z.S.; funding acquisition, Z.S. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


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