**1. Introduction**

The study of *q*-extension of calculus or *q*-analysis motivated the researchers due to its recent use in different applications. In [1,2], Jackson introduced the theory of *q*-calculus. We have seen applications of *q*-analysis in Geometric Function Theory (GFT). They were introduced and applied systematically to the generalized *q*-hypergeometric functions in [3]. Later, Ismail et al. [4] used the *q*-differential operator to examine the geometry of starlike function in *q*-analog. This theory was later extended to the family of *q*-starlike function with some order by Agrawal and Sahoo [5]. Due to this development in function theory, many researchers were motivated, as we have seen by Srivastava in [6]. They added significant contributions, which has slowly made this research area more attractive to forthcoming researchers. We direct the attention of our readers to [7–12] for more information. Moreover, Kanas et al. [13] used Hadamad product to define the *q*-extension of the Ruscheweyh operator. They also discussed in detail some intricate applications of this operator.

Mohammad and Darus [14] conducted an elaborate study of this operator. We have also seen similar work by Mahmood and Sokół [15] and Ahmad et al. [16]. Recently, new thoughts by Maslina in [17] were used to create a novel differential operator called generalized *q*-differential operator with the help of *q*-hypergeometric functions where the authors conducted an in-depth study of applications of this operator. For further information on the extensions of different operators in *q*-analog, we direct the readers to [18–22]. The aim of the present article is to introduce a new integral operator in *q*-analog for multivalent functions using Hadamard product and then study some of its useful applications.

Let <sup>A</sup>*p* (*p* ∈ N = {1, 2, . . .}) contain multivalent functions of all forms *f* that can be defined as holomorphic and/or analytic in any given subset D = {*z* : |*z*| < 1} that is part of a complex plane C which also has the series form shown as:

$$f(z) = z^p + \sum\_{n=1}^{\infty} a\_{n+p} z^{n+p}, \ (z \in \mathbb{D})\,. \tag{1}$$

For any two given functions that are analytic in form *f* and *g* in D, then we can clearly state that *f* is subordinate to *g*, mostly symbolically if it is presented clearly as *f* ≺ *g* or *f* (*z*) ≺ *g* (*z*), if and only if there exists an analytic function *w* with the given properties as *w* (0) = 0 and |*w* (*z*)| < 1 such that *f*(*z*) = *g*(*w*(*z*)) (*z* ∈ <sup>D</sup>). Moreover, if and only if *g* can be seen as univalent in D, then we can clearly have:

$$f(z) \preccurlyeq g(z) \quad (z \in \mathbb{D}) \quad \Longleftrightarrow \quad f(0) = g(0) \quad \text{and} \quad f(\mathbb{D}) \subset \mathcal{g}(\mathbb{D}).$$

For analytic functions *f* of the form Equation (1) and *g* of the form

$$\log(z) = z^p + \sum\_{n=1}^{\infty} b\_{n+p} z^{n+p}, \ (z \in \mathbb{D})\,. \tag{2}$$

the convolution or Hadamard product is defined by

$$(f\*g)(z) = z^p + \sum\_{n=1}^{\infty} a\_{n+p} b\_{n+p} z^{n+p} \ (z \in \mathbb{D})\ .$$

For given *q* ∈ (0, <sup>1</sup>), the derivative in *q*-analog of *f* is given by

$$\mathcal{D}\_q f(z) = \frac{f(z) - f\left(qz\right)}{z\left(1 - q\right)}, \ \left(z \neq 0, \ q \neq 1\right). \tag{3}$$

Making use of Equations (1) and (3), we can easily obtain for *n* ∈ N and *z* ∈ D

$$\mathcal{D}\_q \left\{ \sum\_{n=1}^{\infty} a\_{n+p} z^{n+p} \right\} = \sum\_{n=1}^{\infty} \left[ n + p \right]\_q a\_{n+p} z^{n+p-1},\tag{4}$$

where

$$[n]\_q = \frac{1 - q^n}{1 - q} = 1 + \sum\_{k=1}^{n-1} q^k, \ [0]\_q = 0. \tag{5}$$

For *n* ∈ Z<sup>+</sup> := Z\ {−1, −2, . . .} , the *q*-factorial is given as:

$$\left[n\right]\_q! = \left\{ \begin{array}{c} 1, \ n = 0, \\ \left[1\right]\_q \left[2\right]\_q \dots \left[n\right]\_q, \ n \in \mathbb{N}. \end{array} \right.$$

In addition, with *t* > 0, the *q*-Pochhammer symbol has the form:

$$\begin{aligned} [t]\_{q\mathcal{A}} &= \left( [t]\_q \right)\_n = \begin{cases} 1, \ n = 0, \\\ [t]\_q [t+1]\_q \cdot \cdots [t+n-1]\_q, \ n \in \mathbb{N}, \end{cases} \end{aligned} $$

where [*t*]*q* is given by Equation (5). For *t* > 0, the gamma function in *q*-analog is presented as

$$
\Gamma\_{\emptyset} \left( t + 1 \right) = [t]\_{\emptyset} \Gamma\_{\emptyset} \left( t \right) \text{ and } \Gamma\_{\emptyset} \left( 1 \right) = 1.
$$

We now consider a function

$$\mathcal{F}\_{q,\lambda+p}^{-1}(z) = z^p + \sum\_{n=1}^{\infty} \Psi\_{n-p} a\_{n+p} z^{n+p} \, , \, (\lambda > -p, \, z \in \mathbb{D}) , \tag{6}$$

with

$$\Psi\_{n-p} = \frac{[\lambda + p]\_{q, n+1-p}}{[n+1-p]\_{q}!}.\tag{7}$$

.

We can see that the series given in Equation (6) is absolutely convergen<sup>t</sup> in D. Now, we introduce the integral operator J *λ*+*p*−1 *q* : <sup>A</sup>*p* → <sup>A</sup>*p* by

$$\mathcal{J}\_q^{\lambda+p-1}f(z) = (\mathcal{F}\_{q,\lambda+p}^{-1} \* f)(z) = z^p + \sum\_{n=1}^{\infty} \Psi\_{n-p} a\_{n+p} z^{n+p} \quad (z \in \mathbb{D})\,. \tag{8}$$

where *λ* > <sup>−</sup>*p*. We note that

$$\lim\_{q \to 1^{-}} \mathcal{F}\_{q, \lambda+p}^{-1}(z) = \frac{z^p}{(1-z)^{\lambda+1}} \quad \text{and} \quad \lim\_{q \to 1^{-}} \mathcal{I}\_q^{\lambda+p-1} f(z) = f(z) \* \frac{z^p}{(1-z)^{\lambda+1}}$$

Various integral operators were obtained by putting particular values to the parameters used in the newly defined operator as given by Equation (7):

(i). Making *p* = 1 in our newly defined operator J *λ*+*p*−1 *q f* , we obtain the operator J *λq f* which was introduced by Arif et al. [20] and is given by

$$(\mathcal{J}\_q^\lambda f(z) = (\mathcal{F}\_{q,\lambda+1}^{-1} \* f)(z) = z + \sum\_{n=1}^\infty \Psi\_{n-1} a\_{n+1} z^{n+1} \quad (z \in \mathbb{D})\ .$$

(ii). When *q* → 1<sup>−</sup>, the operator defined in Equation (7) leads to the following well-known Noor integral operator for multivalent functions introduced in [23].

$$\mathcal{J}^{\lambda+p-1}f(z) = (\mathcal{F}\_{\lambda+p}^{-1} \* f)(z) = z^p + \sum\_{n=1}^{\infty} \Psi\_{n-p} a\_{n+p} z^{n+p}, \ (z \in \mathbb{D})\ .$$

(iii). If we set *p* = 1 along with *q* → 1− in Equation (7), then the operator J *λ*+*p*−1 *f* reduced to the following familiar Noor integral operator studied in [24,25].

$$\mathcal{Z}^{\lambda}f(z) = (\mathcal{F}\_{\lambda+1}^{-1} \ast f)(z) = z + \sum\_{n=1}^{\infty} \Psi\_{n-1} a\_{n+1} z^{n+1}, \ (z \in \mathbb{D})\ .$$

For more details on the *q*-analog of differential and integral operators, see [17,26,27].

Motivated from the work in [28–35], we now introduce a subfamily <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*) of <sup>A</sup>*p* by using J *λ*+*p*−1 *q*as follows:

**Definition 1.** *Let f* ∈ <sup>A</sup>*p*. *Then, f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), *if it satisfies the relation*

$$\left| \frac{z^{1-p} \mathcal{D}\_q \mathcal{J}\_q^{\lambda+p-1} f(z) - [p]\_q}{2\beta \left[ z^{1-p} \mathcal{D}\_q \mathcal{J}\_q^{\lambda+p-1} f(z) - a[p]\_q \right] - \left[ z^{1-p} \mathcal{D}\_q \mathcal{J}\_q^{\lambda+p-1} f(z) - [p]\_q \right]} \right| < \mu,\tag{9}$$

*where* 12 *β* < 1, 0 *α* < 12 , 0 < *μ* 1 *and* 0 < *q* < 1.

By varying the parameters values in the class <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), we ge<sup>t</sup> many new classes; we list some of them.


(iv). Further, if we put *p* = 1 and *q* → 1− in <sup>H</sup>*λp*,*<sup>q</sup>* 0, 1, 12 , we have the class <sup>H</sup>*λ*1 0, 1, 12 . Note that we assume throughout our discussion, unless otherwise stated,

$$\frac{1}{2} \le \beta \le 1, \ 0 \le \alpha < 1, \ 0 < \mu \le 1, \ \lambda > -p, \ 0 < q < 1$$

and all coefficients *ak* are positive.

#### **2. The Main Results and Their Consequences**

**Theorem 1.** *If f* ∈ <sup>A</sup>*p has the form of Equation (1) and satisfies the inequality*

$$\sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_q \left( 1 + \mu \left( 2\beta - 1 \right) \right) \left| a\_{n+p} \right| \le 2\mu \beta \left[ p \right]\_q \left( 1 - \alpha \right),\tag{10}$$

 .

*then f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*).

**Proof.** To show that *f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), we just need to prove Equation (9). For this, consider

$$\begin{array}{lcl} L &=& \left| \frac{\frac{1}{\left[p\right]\_{q}} z^{1-p} \mathcal{D}\_{q} \mathcal{J}\_{q}^{\lambda+p-1} f(z) - 1}{2\beta \left[\frac{1}{\left[p\right]\_{q}} z^{1-p} \mathcal{D}\_{q} \mathcal{J}\_{q}^{\lambda+p-1} f(z) - a\right] - \left[\frac{1}{\left[p\right]\_{q}} z^{1-p} \mathcal{D}\_{q} \mathcal{J}\_{q}^{\lambda+p-1} f(z) - 1\right]} \right| \\ &=& \left| \frac{z^{1-p} \mathcal{D}\_{q} \mathcal{J}\_{q}^{\lambda+p-1} f(z) - \left[p\right]\_{q}}{2\beta \left[z^{1-p} \mathcal{D}\_{q} \mathcal{J}\_{q}^{\lambda+p-1} f(z) - a \left[p\right]\_{q}\right] - \left[z^{1-p} \mathcal{D}\_{q} \mathcal{J}\_{q}^{\lambda+p-1} f(z) - \left[p\right]\_{q}\right]} \right| .\end{array}$$

Using Equation (8) with the help of Equations (3) and (4), we can easily obtain

$$\begin{split} L &=& \left| \frac{\sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} a\_{n+p} z^{n}}{2 \beta \left[ p \right]\_{q} (1 - a) + 2 \beta \sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} a\_{n+p} z^{n} - \sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} a\_{n+p} z^{n}} \right| \\ &=& \left| \frac{\sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} a\_{n+p} z^{n}}{2 \beta \left[ p \right]\_{q} (1 - a) + (2\beta - 1) \sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} a\_{n+p} z^{n}} \right| \\ &\leq& \frac{\sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} \left[ a\_{n+p} \right]}{2 \beta \left[ p \right]\_{q} (1 - a) - (2\beta - 1) \sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} \left[ a\_{n+p} \right]} < \mu\_{\prime} \end{split}$$

where we have used the inequality in Equation (10) and this completes the proof.

Making *β* = 12 , *p* = 1 along with *q* → 1<sup>−</sup>, we ge<sup>t</sup> the following result. **Corollary 1.** *If f* ∈ A *and satisfies the inequality*

$$\sum\_{n=1}^{\infty} \Psi\_{n-1} \left( n+1 \right) \left| a\_{n+1} \right| \le \mu \left( 1 - \kappa \right),$$

*then f* ∈ H*<sup>λ</sup> α*, *μ*, 12 .

**Theorem 2.** *If f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*) *has the form of Equation (1), then*

$$r^p - \xi r \le |f(z)| \le r^p + \xi r, |z| = r < 1,$$

*where*

$$\zeta = \frac{2\mu\beta \left[p\right]\_q \left(1 - \alpha\right)}{\left(1 + \mu \left(2\beta - 1\right)\right) \Psi\_{1-p} \left[1 + p\right]\_q}.$$

**Proof.** Consider

$$\begin{aligned} \left| f(z) \right| &= \left| z^p + \sum\_{n=1}^{\infty} a\_{n+p} z^{n+p} \right| \\ &\le \left| z \right|^p + \sum\_{n=1}^{\infty} \left| a\_{n+p} \right| \left| z \right|^{n+p} \end{aligned}$$

$$\begin{aligned} &= \left| r^p + \sum\_{n=1}^{\infty} \left| a\_{n+p} \right| r^{n+p} . \end{aligned}$$

Since 0 < *r* < 1 and *rn*+*<sup>p</sup>* < *r*,

$$\left|f(z)\right| \le r^p + r \sum\_{n=1}^{\infty} \left|a\_{n+p}\right|. \tag{11}$$

.

Similarly,

$$\left|f(z)\right| \ge r^p - r \sum\_{n=1}^{\infty} \left|a\_{n+p}\right|.\tag{12}$$

.

It can easily be seen that

$$\left(\Psi\_{1-p}\left(1+\mu\left(2\beta-1\right)\right)\left[1+p\right]\_q\sum\_{n=1}^{\infty}\left|a\_{n+p}\right|\leq\sum\_{n=1}^{\infty}\Psi\_{n-p}\left(1+\mu\left(2\beta-1\right)\right)\left[n+p\right]\_q\left|a\_{n+p}\right|.\right)$$

By using the relation in Equation (10), we obtain

$$2\left\{\Psi\_{1-p}\left(1+\mu\left(2\beta-1\right)\right)\left[1+p\right]\_q\sum\_{n=1}^{\infty}\left|a\_{n+p}\right|\_q \leq 2\mu\beta\left[p\right]\_q\left(1-a\right)\_n\right\}$$

which gives

$$\sum\_{n=1}^{\infty} \left| a\_{n+p} \right| \quad \le \quad \frac{2\mu\beta[p]\_q(1-\alpha)}{\Psi\_{1-p}(1+\mu(2\beta-1))[1+p]\_q}.$$

Now, by using the above relation in Equations (11) and (12), we obtain the result. Setting *β* = 12, *p* = 1 along with *q* → 1− in the last theorem, we have

**Corollary 2.** *If f* ∈ H*<sup>λ</sup> α*, *μ*, 12 , *then for* |*z*| = *r* < 1

$$\left(1 - \frac{\mu\left(1 - \alpha\right)}{2\left(\lambda + 1\right)}\right)r \le \left|f(z)\right| \le \left(1 + \frac{\mu\left(1 - \alpha\right)}{2\left(\lambda + 1\right)}\right)r.$$

**Theorem 3.** *If f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*) *has the form of Equation (1), then*

$$\begin{array}{rcl} \left[p\right]\_q r^{p-1} - \theta r & \leq & \left[\mathcal{D}\_\theta f(z)\right] & \leq & \left[p\right]\_q r^{p-1} + \theta r \left|z\right| = r < 1, \end{array}$$

*where ϑ* = <sup>2</sup>*μβ*[*p*]*q*(<sup>1</sup>−*<sup>α</sup>*) (<sup>1</sup>+*μ*(2*β*−<sup>1</sup>))<sup>Ψ</sup>1−*<sup>p</sup>*

. **Proof.** By using Equations (3) and (4), we can have

$$\mathcal{D}\_{\emptyset}f(z) = [p]\_{\emptyset}z^{p-1} + \sum\_{n=1}^{\infty} [n+p]\_{\emptyset}a\_{n+p}z^{n+p-1}.$$

Since |*z*|*<sup>p</sup>*−<sup>1</sup> = *rp*−<sup>1</sup> < 1, *rn*+*p*−<sup>1</sup> ≤ *r* and

$$\left|\mathcal{D}\_q f(z)\right| \le \left[p\right]\_q r^{p-1} + r \sum\_{n=1}^{\infty} \left[n+p\right]\_q \left|a\_{n+p}\right|.\tag{13}$$

Similarly,

$$\left| \left| \mathcal{D}\_{q} f(z) \right| \geq \left[ p \right]\_{q} r^{p-1} - r \sum\_{n=1}^{\infty} [n+p]\_{q} \left| a\_{n+p} \right| \,. \tag{14}$$

.

.

*.*

Now, by using Equation (10), we ge<sup>t</sup>

$$\begin{aligned} \left| \Psi\_{1-p} \left( 1 + \mu \left( 2\beta - 1 \right) \right) \sum\_{n=1}^{\infty} \left[ n + p \right]\_q \left| a\_{n+p} \right| \right| &\leq \\ \sum\_{n=1}^{\infty} \left| \Psi\_{n-p} \left[ n + p \right]\_q \left( 1 + \mu \left( 2\beta - 1 \right) \right) \left| a\_{n+p} \right|. \end{aligned}$$

This implies that

$$\sum\_{n=1}^{\infty} \left[n+p\right]\_q \left|a\_{n+p}\right| \quad \overset{<}{\leq} \quad \frac{2\mu\beta[p]\_q(1-\kappa)}{(1+\mu(2\beta-1))\Psi\_{1-p}}$$

Finally, by using above relation in Equations (13) and (14), we have the result.

For *q* → 1<sup>−</sup>, we have the following corollary.

**Corollary 3.** *If f* ∈ H*λp* (*<sup>α</sup>*, *μ*, *β*), *then for* |*z*| = *r* < 1

$$pr^{p-1} - \theta r \le |f'(z)| \le pr^{p-1} + \theta r,$$

*where*

$$\theta = \frac{2\mu\beta \left(1 - \alpha\right)p}{\left(1 + \mu\left(2\beta - 1\right)\right)\Psi\_{1-p}}.$$

**Theorem 4.** *If f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), *then f* ∈ S∗*p* (*δ*) *for* |*z*| < *r*1*, where*

$$r\_1 = \left(\frac{\left(p-\delta\right)\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{n-p}\left[n+p\right]\_q}{2\mu\beta\left(1-\alpha\right)\left(\delta-p\right)\left[p\right]\_q}\right)^{\frac{1}{n}}.$$

**Proof.** Let *f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*). To show that *f* ∈ S∗*p* (*δ*), we have to prove that

$$\left|\frac{zf'(z)-pf(z)}{zf'(z)+(p-2\delta)f(z)}\right|<1.$$

Using Equation (1), we conclude that

$$\sum\_{n=1}^{\infty} \left( \frac{\delta - p}{p - \delta} \right) |a\_{n+p}| \, |z|^n < 1. \tag{15}$$

From Equation (10), it can easily be obtained that

$$\sum\_{n=1}^{\infty} \left( \frac{\Psi\_{n-p} \left[ \mu + p \right]\_q (1 + \mu (2\beta - 1)) \big|}{\left[ p \right]\_q 2\mu \beta (1 - \alpha)} \right) \left| a\_{\mu + p} \right| \quad < \quad 1.1$$

The relation in Equation (15) is true, if the following holds

$$\sum\_{n=1}^{\infty} \left( \frac{\delta - p}{p - \delta} \right) \left| a\_{n+p} \right| \left| z \right|^n < \sum\_{n=1}^{\infty} \left( \frac{\Psi\_{n-p} [n+p]\_q (1 + \mu(2\delta - 1))}{[p]\_q 2\mu \beta (1 - a)} \right) \left| a\_{n+p} \right| \left| z \right| $$

which implies that

$$|z|^{\eta} < \frac{(p-\delta)\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{n-p}\left[n+p\right]\_q}{2\mu\beta\left(1-\alpha\right)\left(\delta-p\right)\left[p\right]\_q}.$$

Therefore,

$$|z| < \left(\frac{(p-\delta)\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{n-p}\left[n+p\right]\_q}{2\mu\beta\left(1-\alpha\right)\left(\delta-p\right)\left[p\right]\_q}\right)^{\frac{1}{n}} = r\_1.$$

Hence, we ge<sup>t</sup> the required result.

Letting *p* = 1 and *q* → 1− in the last theorem, we ge<sup>t</sup> the result below.

**Corollary 4.** *If f* ∈ H*<sup>λ</sup>* (*<sup>α</sup>*, *μ*, *β*), *then f* ∈ S∗ (*δ*) *for* |*z*| < *r*1, *where*

$$r\_1 = \left(\frac{\left(1 - \delta\right)\left(1 + \mu\left(2\beta - 1\right)\right)\left[\left(n + 1\right)\Psi\_{n-1}\right)}{2\mu\beta\left(1 - a\right)\left(\delta - 1\right)}\right)^{\frac{1}{n}}.$$

**Theorem 5.** *If f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), *then f* ∈ C*p* (*δ*) *for* |*z*| < *r*2*, where*

$$r\_2 = \left(\frac{p\left(p-\delta\right)\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{n-p}\left[n+p\right]\_q}{2\mu\beta\left(1-\mu\right)\left[p\right]\_q\left(\delta-p\right)\left(n+p\right)}\right)^{\frac{1}{n}}.$$

**Proof.** Since *f* ∈ C*p* (*δ*),

$$\left|\frac{zf''(z)-(p-1)f'(z)}{zf''(z)+(1-2\delta+p)f'(z)}\right|<1.$$

By using Equation (1) and after some simplifications, we ge<sup>t</sup>

$$\sum\_{n=1}^{\infty} \left( \frac{\left(\delta - p\right)\left(n + p\right)}{p\left(p - \delta\right)} \right) \left| a\_{n+p} \right| \left| z \right|^n < 1. \tag{16}$$

Now, from Equation (10), we can easily obtain that

$$\sum\_{n=1}^{\infty} \left( \frac{(1 + \mu(2\beta - 1))\Psi\_{n-p}[n+p]\_q}{2\mu\beta(1-\alpha)|p|\_q} \right) \left| a\_{n+p} \right| \quad < \quad 1.1$$

> The relation in Equation (16) is true if

$$\sum\_{n=1}^{\infty} \left( \frac{(\delta - p)(n + p)}{p(p - \delta)} \right) \left| a\_{\mathcal{H} + p} \right| \left| z \right|^n \quad \prec \sum\_{n=1}^{\infty} \left( \frac{(1 + \mu(2\delta - 1)) \Psi\_{n - p}[n + p]\_q}{2\mu \delta (1 - \alpha) |p|\_q} \right) \left| a\_{\mathcal{H} + p} \right| \; \mathcal{F}$$

which gives

$$\left|z\right|^{n} < \left(\frac{p\left(p-\delta\right)\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{n-p}\left[n+p\right]\_{q}}{2\mu\beta\left(1-\mu\right)\left[p\right]\_{q}\left(\delta-p\right)\left(n+p\right)}\right).$$

Hence,

$$|z| < \left(\frac{p\left(p-\delta\right)\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{n-p}\left[n+p\right]\_q}{2\mu\beta\left(1-\alpha\right)\left[p\right]\_q\left(\delta-p\right)\left(n+p\right)}\right)^{\frac{1}{n}} = r\_2.$$

Thus, we obtain the required result.

Substituting *p* = 1 and taking *q* → 1− in the last theorem, we ge<sup>t</sup> the corollary below.

**Corollary 5.** *If f* ∈ H*<sup>λ</sup>* (*<sup>α</sup>*, *μ*, *β*), *then f* ∈ C (*δ*) *for* |*z*| < *r*2, *where*

$$r\_2 = \left(\frac{\left(1-\delta\right)\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{n-1}}{2\mu\beta\left(1-\alpha\right)\left(\delta-1\right)}\right)^{\frac{1}{n}}.$$

**Theorem 6.** *Let fp* (*z*) = *zp and*

$$f\_k(z) = z^p + \frac{2\mu\beta\left(1 - a\right)|p|\_q}{\left(1 + \mu\left(2\beta - 1\right)\right)\Psi\_{k-2p}\left[k\right]\_q a\_k} z^k, \ (k \ge n + p)\,. \tag{17}$$

*Then, f* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), *if and only if it can be expressed in the form*

$$f\left(z\right) = \lambda\_p z^p + \sum\_{k=n+p}^{\infty} \lambda\_k f\_k\left(z\right),\tag{18}$$

.

*where λp* ≥ 0*, λk* ≥ 0, *k* ≥ *n* + *p and λp* + ∞ ∑ *k*=*n*+*p λk* = 1.

**Proof.** We suppose that *f* can be written of the form of Equation (18), thus

$$\begin{split} f\left(z\right) &=& \lambda\_p z^p + \sum\_{k=n+p}^{\infty} \lambda\_k f\_k\left(z\right), \\ &=& \lambda\_p z^p + \sum\_{k=n+p}^{\infty} \lambda\_k \left(z^p + \frac{2\mu\beta\left(1-a\right)\left[p\right]\_q}{\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{k-2p}\left[k\right]\_q a\_k} z^k\right), \\ &=& z^p + \sum\_{k=n+p}^{\infty} \frac{2\mu\beta\left(1-a\right)\left[p\right]\_q \lambda\_k}{\left(1+\mu\left(2\beta-1\right)\right)\Psi\_{k-2p}\left[k\right]\_q a\_k} z^k. \end{split}$$

This implies that

$$\begin{split} \sum\_{k=n+p}^{\infty} \frac{(1+\mu(2\beta-1))\Psi\_{k-2p}[k]\_q a\_k}{2\mu\beta(1-a)|p|\_q} &\times \frac{2\mu\beta(1-a)|p|\_q}{(1+\mu(2\beta-1))\Psi\_{k-2p}[k]\_q a\_k} \lambda\_{k'} \\ &= \sum\_{k=n+p}^{\infty} \lambda\_{k'} \\ &= 1 - \lambda\_p \le 1. \end{split}$$

Conversely, we suppose that *fn* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*). Then, by using Equation (10), we have

$$|a\_k| \le \frac{2\mu\beta\left(1 - a\right) \left[p\right]\_q}{\left(1 + \mu\left(2\beta - 1\right)\right)\Psi\_{k-2p}\left[k\right]\_q}, \ k \ge p + n.$$

By setting

$$
\lambda\_k = \frac{\left(1 + \mu\left(2\beta - 1\right)\right) \Psi\_{k-2p}\left[k\right]\_q}{2\mu\beta \left(1 - \alpha\right) \left[p\right]\_q} a\_{k\nu}
$$

then

$$\begin{split} f\left(z\right) &= \quad z^{p} + \sum\_{k=n+p}^{\infty} a\_{k} z^{k} \\ &= \quad z^{p} + \sum\_{k=n+p}^{\infty} \frac{2\mu\beta\left(1-\alpha\right) \left[p\right]\_{q}}{\left(1 + \mu\left(2\beta - 1\right)\right) \Psi\_{k-2p}\left[k\right]\_{q} a\_{k}} \lambda\_{k} z^{k} \\ &= \quad z^{p} + \sum\_{k=n+p}^{\infty} \left[z^{p} - f\_{k}\left(z\right)\right] \lambda\_{k} \\ &= \quad \left(1 - \sum\_{k=n+p}^{\infty} \lambda\_{k}\right) z^{p} + \sum\_{k=n+p}^{\infty} \lambda\_{k} f\_{k}\left(z\right) \\ &= \quad \lambda\_{p} z^{p} + \sum\_{k=n+p}^{\infty} \lambda\_{k} f\_{k}\left(z\right) .\end{split}$$

This complete the result.

Putting *p* = 1, *β* = 12 and *q* → 1− in the above theorem, we obtain the upcoming corollary. **Corollary 6.** *Let f* (*z*) = *z and*

$$f\_k(z) = z + \frac{\mu \left(1 - \kappa\right)}{k \Psi\_{k-2} a\_k} z^k, \ (k \ge n + 1) \,. \tag{19}$$

*Then, f* ∈ H∗*p α*, *μ*, 12 , *if and only if it can be expressed in the form*

$$f\left(z\right) = \lambda z + \sum\_{k=n+1}^{\infty} \lambda\_k f\_k\left(z\right),\tag{20}$$

*where λ* ≥ 0, *λk* ≥ 0, *k* ≥ *n* + 1 *and λ*+ ∞ ∑ *k*=*n*+1 *λk* = 1.

**Theorem 7.** *If f* , *g* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), *then* (*f* ∗ *g*) ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*), *where*

$$d \ge \mu \left( \frac{2l\beta \left(1 - a\right) \left[p\right]\_q}{\Psi\_{n-p} \left[n + p\right]\_q} - l \left(2\beta - 1\right) \right). \tag{21}$$

**Proof.** We have to find largest *μ* such that

$$\sum\_{n=1}^{\infty} \frac{(1 + \mu \left(2\beta - 1\right)) \Psi\_{n-p} \left[n + p\right]\_q}{2\mu \beta \left(1 - \kappa\right) \left[p\right]\_q} a\_{n+p} b\_{n+p} \le 1.$$

Let *f* , *g* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*). Then, using Equation (10), we obtain

$$\sum\_{n=1}^{\infty} \frac{\left(1 + l\left(2\beta - 1\right)\right) \Psi\_{n-p}\left[n+p\right]\_q}{2l\beta \left(1 - \alpha\right) \left[p\right]\_q} a\_{n+p} \le 1\tag{22}$$

and

$$\sum\_{n=1}^{\infty} \frac{\left(1 + l\left(2\beta - 1\right)\right) \Psi\_{n-p}\left[n+p\right]\_q}{2l\beta \left(1 - \alpha\right) \left[p\right]\_q} b\_{n+p} \le 1. \tag{23}$$

By using Cauchy–Schwarz inequality, we have

$$\sum\_{n=1}^{\infty} \frac{\left(1 + l\left(2\beta - 1\right)\right) \Psi\_{n-p}\left[n+p\right]\_q}{2l\beta\left(1-a\right)\left[p\right]\_q} \sqrt{a\_{n+p} b\_{n+p}} \le 1. \tag{24}$$

Thus, we have to show that

$$\begin{aligned} &\frac{(1+l\left(2\beta-1\right))\,\Psi\_{n-p}\left[n+p\right]\_q}{2l\beta\left(1-\alpha\right)\left[p\right]\_q}a\_{n+p}b\_{n+p} \\ &\leq \quad \frac{(1+\mu\left(2\beta-1\right))\,\Psi\_{n-p}\left[n+p\right]\_q}{2\mu\beta\left(1-\alpha\right)\left[p\right]\_q}\sqrt{a\_{n+p}b\_{n+p}}\end{aligned}$$

that is

$$\sqrt{a\_{n+p}b\_{n+p}} \le \frac{l}{\mu} \frac{(1+\mu\left(2\beta -1\right))}{(1+l\left(2\beta -1\right))}.\tag{25}$$

In addition, from Equation (24), we can write

$$\sqrt{a\_{n+p}b\_{n+p}} \le \frac{2l\beta \left(1 - \alpha\right) \left[p\right]\_q}{\left(1 + l\left(2\beta - 1\right)\right) \Psi\_{n-p}\left[n+p\right]\_q}.\tag{26}$$

Consequently, we have to show that

$$\frac{2l\beta\left(1-\alpha\right)\left[p\right]\_q}{\left(1+l\left(2\beta-1\right)\right)\Psi\_{n-p}\left[n+p\right]\_q} \le \frac{l\left(1+\mu\left(2\beta-1\right)\right)}{\mu\left(1+l\left(2\beta-1\right)\right)}.$$

By simple calculation, we ge<sup>t</sup>

$$d \ge \mu \left( \frac{2l\beta \left(1 - \alpha\right) \left[p\right]\_q}{\Psi\_{n-p}^\* \left[n + p\right]\_q} - l \left(2\beta - 1\right) \right),$$

which completes the required result.

**Theorem 8.** *Let f*1 *and f*2 *be in the class* <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*). *Then, the weighted mean wq of f*1 *and f*2 *is also in the class* <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*).

**Proof.** From the definition of weighted mean, we have

$$\begin{aligned} w\_q &= \quad \frac{1}{2} \left[ (1-q) \, f\_1 \left( z \right) + (1+q) \, f\_2 \left( z \right) \right] \\ &= \quad \left[ \left( 1-q \right) \left( z^p + \sum\_{n=1}^{\infty} a\_{n+p} z^{n+p} \right) + (1+q) \left( z^p + \sum\_{n=1}^{\infty} b\_{n+p} z^{n+p} \right) \right] \\ &= \quad z^p + \sum\_{n=1}^{\infty} \frac{1}{2} \left[ \left( 1-q \right) a\_{n+p} + \left( 1+q \right) b\_{n+p} \right] z^{n+p} .\end{aligned}$$

> Now, using Equation (10), we have

$$\sum\_{n=1}^{n} \Psi\_{n-p} \left[ n + p \right]\_q \left( 1 + \mu \left( 2\beta - 1 \right) \right) a\_{n+p} \le 2\mu \beta \left[ p \right]\_q \left( 1 - \alpha \right)^p$$

and

$$\sum\_{n=1}^{n} \Psi\_{n-p} \left[ n + p \right]\_q \left( 1 + \mu \left( 1 - 2\beta \right) \right) b\_{n+p} \le 2\mu \beta \left[ p \right]\_q \left( 1 - \kappa \right).$$

Consider

$$\begin{split} & \quad \left[ \sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} \left( 1 + \mu \left( 2\beta - 1 \right) \right) \right] \left[ \frac{1}{2} \left[ \left( 1 - q \right) a\_{n+p} + \left( 1 + q \right) b\_{n+p} \right] \right] \\ &= \quad \frac{1}{2} \left( 1 - q \right) \sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} \left( 1 + \mu \left( 2\beta - 1 \right) \right) a\_{n+p} \\ & \quad + \frac{1}{2} \left( 1 + q \right) \sum\_{n=1}^{\infty} \Psi\_{n-p} \left[ n + p \right]\_{q} \left( 1 + \mu \left( 2\beta - 1 \right) \right) b\_{n+p} \\ & \leq \quad \frac{1}{2} \left( 1 - q \right) 2\mu \beta \left[ p \right]\_{q} \left( 1 - a \right) + \frac{1}{2} \left( 1 + q \right) 2\mu \beta \left[ p \right]\_{q} \left( 1 - a \right) . \end{split}$$

This shows that *wq* ∈ <sup>H</sup>*λp*,*<sup>q</sup>* (*<sup>α</sup>*, *μ*, *β*).
