Properties of a New Subclass of Analytic Functions With Negative Coefficients Defined by Using the Q-Derivative

The quantum calculus or q–calculus has called the attention of many great researchers from Geometric function theory field due to its numerous applications in mathematics and physics. In 1999 was defined the first q-analogue of a starlike function by Ismail et al. [1]. In fact, many results obtained for univalent functions can be extended by using q-analogues functions. Recently, analytic functions with negative coefficient were studied in papers [5, 8, 9, 10].

Let U = {z : |z| < 1} be the open unit disk of the complex plane and let 𝒜 represent the class of all functions of the form (1.1) $f (z) = z + \sum_{n = 2}^{\infty} a_{n} z^{n}, z \in U .$ f(z) = z + \sum\limits_{n = 2}^\infty {a_n}{z^n},\;\;z \in U.

For 0 < q < 1, the q-derivative of a function f ∈ 𝒜 is defined by (see [2]) (1.2) $\begin{matrix} d_{q} f (z) = \frac{f (qz) - f (z)}{(q - 1) z}, & (z \neq 0), \end{matrix}$ \matrix{ {{d_q}f(z) = {{f(qz) - f(z)} \over {(q - 1)z}},} & {(z \ne 0),}} with d_q f (0) = f′ (0).

Motivated by the aforementionated works, we define the following class of functions associated with Janowski functions:

Definition 1.1

For 0 ≤ μ ≤ 1, k ≥ 0 and −1 ≤ B < A ≤ 1, we let ST (q, μ,k;A,B) be the subclass of 𝒜 consisting of functions of the form (1.1) and satisfying the condition (1.3) $Re {\frac{(B - 1) {zd}_{q} f (z) / [(1 - μ) z + μ f (z)] - (A - 1)}{(B + 1) {zd}_{q} f (z) / [(1 - μ) z + μ f (z)] - (A + 1)}} >$ {\rm Re} \left\{ {{{(B - 1)z{d_q}f(z)/[(1 - \mu )z + \mu f(z)] - (A - 1)} \over {(B + 1)z{d_q}f(z)/[(1 - \mu )z + \mu f(z)] - (A + 1)}}} \right\} > (1.4) $k | \frac{(B - 1) {zd}_{q} f (z) / [(1 - μ) z + μ f (z)] - (A - 1)}{(B + 1) {zd}_{q} f (z) / [(1 - μ) z + μ f (z)] - (A + 1)} - 1 |, z \in U .$ k\left| {{{(B - 1)z{d_q}f(z)/[(1 - \mu )z + \mu f(z)] - (A - 1)} \over {(B + 1)z{d_q}f(z)/[(1 - \mu )z + \mu f(z)] - (A + 1)}} - 1} \right|,{\kern 1pt} \,z \in U.

Let 𝒯 denote the subclass of analytic functions f ∈ 𝒜 of the form (1.5) $f (z) = z - \sum_{n = 2}^{\infty} a_{n} z^{n}, a_{n} \geq 0 .$ f(z) = z - \sum\limits_{n = 2}^\infty {a_n}{z^n},\;\;{a_n} \ge 0. Further, let 𝒯ST (q, μ,k;A,B) ≡ ST (q, μ,k;A,B) ∩ 𝒯.

Also, we remark that ST (q, μ,k;A,B) reduces to the following known classes: (i)

In case A = 1 − 2α, 0 ≤ α < 1, B = −1, μ = 1 and q → 1⁻, we obtain the class S_p(k,α) of k–uniformly starlike functions of order α(see [4] );

(ii)

In case A = 1, B = −1, μ = 1, k = 1 and q → 1⁻, we get the class S_p of uniformly starlike functions (see [6]);

(iii)

In case μ = 1 and q → 1⁻, we obtain the class k − ST [A,B] which was introduced and studied by Noor and Malik (see [3]);

(iv)

In case A = 1 − 2α, 0 ≤ α < 1, B = −1, μ = 1, k = 0 and q → 1⁻, we get the class of starlike functions of order α denoted S^* (α);

(v)

In case A = 1, B = −1, μ = 0, k = 0 and q → 1⁻, we obtain the class R of functions whose derivative has positive real part.

Unless otherwise mentioned, we assume throughout this paper that 0 ≤ μ ≤ 1, k ≥ 0 and −1 ≤ B < A ≤ 1.

Coefficient Estimates

We begin with a result that provides coefficient inequalities for functions in the class ST (q, μ,k;A,B).

Theorem 2.1

A function f ∈ 𝒜 of the form (1.1) is in the class ST (q, μ,k;A,B) if(2.1) $\sum_{n = 2}^{\infty} {2 (k + 1) ({[n]}_{q} - μ) + | {[n]}_{q} (B + 1) - μ (A + 1) |} \cdot | a_{n} | \leq | B - A | .$ \sum\limits_{n = 2}^\infty \{ 2(k + 1)({[n]_q} - \mu ) + |{[n]_q}(B + 1) - \mu (A + 1)|\} \cdot |{a_n}| \le |B - A|.

Proof

It is suffices to prove that (2.2) $k | g (z) - 1 | - Re [g (z) - 1] < 1, z \in U,$ k|g(z) - 1| - {\rm Re} [g(z) - 1] < 1,\quad z \in U, where the function g is defined by (2.3) $g (z) : = \frac{(B - 1) {zd}_{q} f (z) / [(1 - μ) z + μ f (z)] - (A - 1)}{(B + 1) {zd}_{q} f (z) / [(1 - μ) z + μ f (z)] - (A + 1)}, z \in U .$ g(z): = {{(B - 1)z{d_q}f(z)/[(1 - \mu )z + \mu f(z)] - (A - 1)} \over {(B + 1)z{d_q}f(z)/[(1 - \mu )z + \mu f(z)] - (A + 1)}},\quad z \in U.

First of all, (2.4) $k | g (z) - 1 | - Re [g (z) - 1] \leq (k + 1) | g (z) - 1 |, z \in U .$ k|g(z) - 1| - {\rm Re} [g(z) - 1] \le (k + 1)|g(z) - 1|,\quad z \in U.

Because (2.5) $\begin{array}{l} (k + 1) | g (z) - 1 | & = \frac{2 (k + 1) | \sum_{n = 2}^{\infty} (μ - {[n]}_{q}) a_{n} z^{n - 1} |}{| (B - A) + \sum_{n = 2}^{\infty} {(B + 1) {[n]}_{q} - μ (A + 1)} a_{n} z^{n - 1} |} \\ \leq \frac{2 (k + 1) \sum_{n = 2}^{\infty} (μ - {[n]}_{q}) | a_{n} |}{| B - A | - \sum_{n = 2}^{\infty} {(B + 1) {[n]}_{q} - μ (A + 1)} | a_{n} |} \end{array}$ \matrix{ {(k + 1)|g(z) - 1|} \hfill & { = {{2(k + 1)|\sum\limits_{n = 2}^\infty (\mu - {{[n]}_q}){a_n}{z^{n - 1}}|} \over {|(B - A) + \sum\limits_{n = 2}^\infty \{ (B + 1){{[n]}_q} - \mu (A + 1)\} {a_n}{z^{n - 1}}|}}} \hfill \cr {} \hfill & { \le {{2(k + 1)\sum\limits_{n = 2}^\infty (\mu - {{[n]}_q})|{a_n}|} \over {|B - A| - \sum\limits_{n = 2}^\infty \{ (B + 1){{[n]}_q} - \mu (A + 1)\} |{a_n}|}}}} and (2.1) take place, then (k + 1)|g(z) − 1| is bounded above by 1. Hence f ∈ ST (q, μ,k;A,B) and the theorem is proved.

Corollary 2.1

(see [3], Theorem 2.1) Let function f ∈ 𝒜 be of the form (1.1). If(2.6) $\sum_{n = 2}^{\infty} {2 (k + 1) (n - 1) + | n (B + 1) - (A + 1) |} \cdot | a_{n} | \leq | B - A |,$ \sum\limits_{n = 2}^\infty \{ 2(k + 1)(n - 1) + |n(B + 1) - (A + 1)|\} \cdot |{a_n}| \le |B - A|,then f ∈ k − ST [A,B].

Corollary 2.2

(see [7], Theorem 1) Let function f ∈ 𝒜 be of the form (1.1). If(2.7) $\sum_{n = 2}^{\infty} (n - α) \cdot | a_{n} | \leq 1 - α,$ \sum\limits_{n = 2}^\infty (n - \alpha ) \cdot |{a_n}| \le 1 - \alpha ,then function f is in the class S^* (α).

For f ∈ 𝒯ST (q, μ,k;A,B) the converse of Theorem 2.1 is also true.

Theorem 2.2

A function f ∈ 𝒯 given by (1.5) is in the class 𝒯ST (q, μ,k;A,B) if and only if(2.8) $\sum_{n = 2}^{\infty} {2 (k + 1) ({[n]}_{q} - μ) + | {[n]}_{q} (B + 1) - μ (A + 1) |} \cdot a_{n} \leq | B - A | .$ \sum\limits_{n = 2}^\infty \{ 2(k + 1)({[n]_q} - \mu ) + |{[n]_q}(B + 1) - \mu (A + 1)|\} \cdot {a_n} \le |B - A|.The result is sharp with the extremal function f given by(2.9) $f (z) = z - \frac{| B - A |}{2 (k + 1) ({[n]}_{q} - μ) + | {[n]}_{q} (B + 1) - μ (A + 1) |} z^{n}, z \in U .$ f(z) = z - {{|B - A|} \over {2(k + 1)({{[n]}_q} - \mu ) + |{{[n]}_q}(B + 1) - \mu (A + 1)|}}{z^n},\quad z \in U.

Proof

In view of Theorem 2.1, we need only to prove that (2.8) holds if f ∈ 𝒯ST (q, μ,k;A,B). Conversely, assuming that f ∈ 𝒯ST (q, μ,k;A,B) and z is real, we find that g(z) ≥ k|g(z) − 1|, where g is given in (2.3). Therefore, (2.10) $\begin{array}{l} \frac{(B - A) - \sum_{n = 2}^{\infty} {(B - 1) {[n]}_{q} - μ (A - 1)} a_{n} z^{n - 1}}{(B - A) - \sum_{n = 2}^{\infty} {(B + 1) {[n]}_{q} - μ (A + 1)} a_{n} z^{n - 1}} \geq \\ 2 k | \frac{\sum_{n = 2}^{\infty} ({[n]}_{q} - μ) a_{n} z^{n - 1}}{(B - A) - \sum_{n = 2}^{\infty} {(B + 1) {[n]}_{q} - μ (A + 1)} a_{n} z^{n - 1}} | . \end{array}$ \matrix{ {{{(B - A) - \Sigma_{n = 2}^\infty \{ (B - 1){{[n]}_q} - \mu (A - 1)\} {a_n}{z^{n - 1}}} \over {(B - A) - \Sigma_{n = 2}^\infty \{ (B + 1){{[n]}_q} - \mu (A + 1)\} {a_n}{z^{n - 1}}}} \ge } \hfill \cr {2k\left| {{{\Sigma_{n = 2}^\infty ({{[n]}_q} - \mu ){a_n}{z^{n - 1}}} \over {(B - A) - \Sigma_{n = 2}^\infty \{ (B + 1){{[n]}_q} - \mu (A + 1)\} {a_n}{z^{n - 1}}}}} \right|.}} Next, letting z → 1⁻ through real values, we obtain the inequality (2.8). Also, the equality in (2.8) take place for the function f given in (2.9).

Corollary 2.3

Let function f be of the form (1.5). Then f ∈ k − ST [A,B] ∩ 𝒯 if and only if(2.11) $\sum_{n = 2}^{\infty} {2 (k + 1) (n - 1) + | n (B + 1) - (A + 1) |} \cdot a_{n} \leq | B - A | .$ \sum\limits_{n = 2}^\infty \{ 2(k + 1)(n - 1) + |n(B + 1) - (A + 1)|\} \cdot {a_n} \le |B - A|.

Corollary 2.4

(see [7], Theorem 2) Let function f ∈ 𝒯 be of the form (1.5). Then f is a starlike function of order α if and only if(2.12) $\sum_{n = 2}^{\infty} (n - α) \cdot a_{n} \leq 1 - α .$ \sum\limits_{n = 2}^\infty (n - \alpha ) \cdot {a_n} \le 1 - \alpha .

Distortion Theorems

Theorem 3.1

Let the function f of the form (1.5) be in the class 𝒯ST (q, μ,k;A,B). Then, for |z| = r, (3.1) $| f (z) | \leq r + \frac{| B - A |}{2 (k + 1) ({[2]}_{q} - μ) + | {[2]}_{q} (B + 1) - μ (A + 1) |} r^{2}$ |f(z)| \le r + {{|B - A|} \over {2(k + 1)({{[2]}_q} - \mu ) + |{{[2]}_q}(B + 1) - \mu (A + 1)|}}{r^2}and(3.2) $| f (z) | \geq r - \frac{| B - A |}{2 (k + 1) ({[2]}_{q} - μ) + | {[2]}_{q} (B + 1) - μ (A + 1) |} r^{2},$ |f(z)| \ge r - {{|B - A|} \over {2(k + 1)({{[2]}_q} - \mu ) + |{{[2]}_q}(B + 1) - \mu (A + 1)|}}{r^2},with equality for(3.3) $f (z) = z - \frac{| B - A |}{2 (k + 1) ({[2]}_{q} - μ) + | {[2]}_{q} (B + 1) - μ (A + 1) |} z^{2} .$ f(z) = z - {{|B - A|} \over {2(k + 1)({{[2]}_q} - \mu ) + |{{[2]}_q}(B + 1) - \mu (A + 1)|}}{z^2}.

Proof

By using the hypothesis and the triangle inequality, we find that (3.4) $r - r^{2} \sum_{n = 2}^{\infty} a_{n} \leq | f (z) | \leq r + r^{2} \sum_{n = 2}^{\infty} a_{n},$ r - {r^2}\sum\limits_{n = 2}^\infty {a_n} \le |f(z)| \le r + {r^2}\sum\limits_{n = 2}^\infty {a_n}, which, in conjunction with (2.8) gives us the inequalities (3.1) and (3.2).

Corollary 3.1

Let the function f given by (1.5) be in the class k − ST [A,B] ∩ 𝒯. Then, for |z| = r, $r - \frac{| B - A |}{2 (k + 1) + | 2 B - A + 1 |} r^{2} \leq | f (z) | \leq r + \frac{| B - A |}{2 (k + 1) + | 2 B - A + 1 |} r^{2},$ r - {{|B - A|} \over {2(k + 1) + |2B - A + 1|}}{r^2} \le |f(z)| \le r + {{|B - A|} \over {2(k + 1) + |2B - A + 1|}}{r^2},with equality for(3.5) $f (z) = z - \frac{| B - A |}{2 (k + 1) + | 2 B - A + 1 |} z^{2} .$ f(z) = z - {{|B - A|} \over {2(k + 1) + |2B - A + 1|}}{z^2}.

Corollary 3.2

(see [7], Theorem 4) Let the function f ∈ 𝒯 given by (1.5) be in the class S^* (α) ∩ 𝒯. Then, for |z| = r, $r - \frac{1 - α}{2 - α} r^{2} \leq | f (z) | \leq r + \frac{1 - α}{2 - α} r^{2},$ r - {{1 - \alpha } \over {2 - \alpha }}{r^2} \le |f(z)| \le r + {{1 - \alpha } \over {2 - \alpha }}{r^2},with equality for(3.6) $f (z) = z - \frac{1 - α}{2 - α} z^{2} .$ f(z) = z - {{1 - \alpha } \over {2 - \alpha }}{z^2}.

The next theorem can be proven by employing similar techniques as in the demonstration of Theorem 3.1, so we will omit the details of our proof.

Theorem 3.2

Let the function f of the form (1.5) be in the class 𝒯ST (q, μ,k;A,B). Then, for |z| = r, (3.7) $| f^{'} (z) | \leq 1 + \frac{2 | B - A |}{2 (k + 1) ({[2]}_{q} - μ) + | {[2]}_{q} (B + 1) - μ (A + 1) |} r$ |f^{'}(z)| \le 1 + {{2|B - A|} \over {2(k + 1)({{[2]}_q} - \mu ) + |{{[2]}_q}(B + 1) - \mu (A + 1)|}}rand(3.8) $| f^{'} (z) | \geq 1 - \frac{2 | B - A |}{2 (k + 1) ({[2]}_{q} - μ) + | {[2]}_{q} (B + 1) - μ (A + 1) |} r,$ |f^{'}(z)| \ge 1 - {{2|B - A|} \over {2(k + 1)({{[2]}_q} - \mu ) + |{{[2]}_q}(B + 1) - \mu (A + 1)|}}r,with equality for f given by (3.3).

Corollary 3.3

Let the function f given by (1.5) be in the class k − ST [A,B] ∩ 𝒯. Then, for |z| = r, $1 - \frac{2 | B - A |}{2 (k + 1) + | 2 B - A + 1 |} r \leq | f^{'} (z) | \leq 1 + \frac{2 | B - A |}{2 (k + 1) + | 2 B - A + 1 |} r,$ 1 - {{2|B - A|} \over {2(k + 1) + |2B - A + 1|}}r \le |f^{'}(z)| \le 1 + {{2|B - A|} \over {2(k + 1) + |2B - A + 1|}}r,with equality for f given by (3.5).

Corollary 3.4

(see [7], Theorem 4) Let the function f ∈ 𝒯 given by (1.5) be in the class S^* (α) ∩ 𝒯. Then, for |z| = r, $1 - \frac{2 (1 - α)}{2 - α)} r \leq | f^{'} (z) | \leq 1 + \frac{2 (1 - α)}{2 - α} r,$ 1 - {{2(1 - \alpha )} \over {2 - \alpha )}}r \le |f^{'}(z)| \le 1 + {{2(1 - \alpha )} \over {2 - \alpha }}r,with equality for f given by (3.6).

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Properties of a New Subclass of Analytic Functions With Negative Coefficients Defined by Using the Q-Derivative

Published Online: Mar 31, 2020

Page range: 303 - 308

Received: Jun 15, 2019

Accepted: Oct 15, 2019

DOI: https://doi.org/10.2478/amns.2020.1.00028

Keywordsanalytic functions, Janowski functions, negative coefficients, q-calculus, q-differential operator

© 2020 Roberta Bucur et al., published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 International License.

Keywords
analytic functions, Janowski functions, negative coefficients, q-calculus, q-differential operator